Sequence and Series for JEE: AP, GP, HP, Special Sums and Differences

A sequence is a list of numbers in a definite order: a first term, a second term, a third term, and so on. When a rule produces each term from its position, the whole list can be described in one line, and its sum can often be found without adding term by term. Arithmetic and geometric progressions are the two simplest rules, and almost every series asked in JEE is built from them, from the sums of $n$, $n^2$ and $n^3$, or from a term that splits into a difference.

This note covers sequences and the relation between $a_n$ and $S_n$, the arithmetic progression with its sum, properties and the choice of terms, the geometric progression with finite and infinite sums, problems that mix an AP with a GP, the harmonic progression, the three means and the insertion of means, the arithmetico-geometric progression, the special sums $\sum n$, $\sum n^2$, $\sum n^3$, the method of differences, sequences given by a recurrence, and bounding a sum. The inequality $A\ge G\ge H$ and its use for maxima and minima is developed fully in the Relation Between AM, GM and HM note; here it gets one short section. Each section ends with worked examples from JEE Main, JEE Advanced and IIT JEE papers, graded from easy to hard.

Sequences, series and the sum $S_n$

A sequence is a function from the natural numbers to the real numbers. Its value at $n$ is the $n$th term, written $a_n$ or $T_n$:

$$a_1,\ a_2,\ a_3,\ \dots,\ a_n,\ \dots$$

A sequence with a last term is finite; one that never stops is infinite. The expression obtained by adding the terms is a series, and the sum of the first $n$ terms is

$$S_n=a_1+a_2+\dots+a_n=\sum_{k=1}^{n}a_k$$

From $S_n$ back to $a_n$

$S_n$ and $S_{n-1}$ differ only by the last term, so

$$a_n=S_n-S_{n-1},\quad n\ge2$$ $$a_1=S_1$$

Rules for $\Sigma$

For constants $c$ and any two sequences,

$$\sum_{k=1}^{n}(c\,a_k+b_k)=c\sum_{k=1}^{n}a_k+\sum_{k=1}^{n}b_k$$ $$\sum_{k=1}^{n}c=nc$$

But a sum of products is not the product of sums: $\sum a_kb_k\ne\left(\sum a_k\right)\left(\sum b_k\right)$ in general.

Examples

Example 1 (JEE Main 2026, 23 Jan Shift 2). Let $\displaystyle\sum_{k=1}^{n}a_k=\alpha n^2+\beta n$. If $a_{10}=59$ and $a_6=7a_1$, then $\alpha+\beta$ is equal to

▸ SHOW SOLUTION

Step 1. Find $a_n$ from $S_n$

$$a_n=\alpha n^2+\beta n-\alpha(n-1)^2-\beta(n-1)$$$$a_n=\alpha(2n-1)+\beta$$

Here $S_0=0$, so this also gives $a_1=\alpha+\beta$ correctly.

Step 2. Use the two conditions

$a_{10}=19\alpha+\beta=59$. Also $a_6=11\alpha+\beta$ and $7a_1=7\alpha+7\beta$, so

$$11\alpha+\beta=7\alpha+7\beta\Rightarrow\beta=\frac{2\alpha}{3}$$

Step 3. Solve

$$19\alpha+\frac{2\alpha}{3}=59\Rightarrow\alpha=3,\ \beta=2$$

Answer: $\alpha+\beta=5$, option (3)

Example 2 (JEE Main 2026, 2 April Shift 1). If $\displaystyle\sum_{k=1}^{n}a_k=6n^3$, then $\displaystyle\sum_{k=1}^{6}\left(\frac{a_{k+1}-a_k}{36}\right)^2$ is equal to ______.

▸ SHOW SOLUTION

Step 1. Find $a_n$

$$a_n=6n^3-6(n-1)^3=18n^2-18n+6$$

At $n=1$ this gives $6=S_1$, so it holds for every $n$.

Step 2. Difference of consecutive terms

$$a_{k+1}-a_k=18(2k+1)-18=36k$$

Step 3. Sum

$$\sum_{k=1}^{6}k^2=\frac{6\cdot7\cdot13}{6}=91$$

Answer: $91$

Example 3 (IITianForum). A sequence of positive numbers $t_1,t_2,t_3,\dots$ has $t_1=1$, and the sum $S_n$ of its first $n$ terms satisfies $S_n=\dfrac12\left(t_n+\dfrac{1}{t_n}\right)$ for every $n$. Then $S_{25}-t_{25}$ is equal to

▸ SHOW SOLUTION

Step 1. Replace $t_n$ by $S_n-S_{n-1}$

For $n\ge2$, $t_n=S_n-S_{n-1}$, so $2S_n=t_n+\dfrac{1}{t_n}$ becomes

$$2S_n-(S_n-S_{n-1})=\frac{1}{S_n-S_{n-1}}$$$$(S_n+S_{n-1})(S_n-S_{n-1})=1$$

Step 2. The squares form an AP

$$S_n^2-S_{n-1}^2=1,\qquad S_1^2=1$$

So $S_n^2=n$ and $S_n=\sqrt n$, since every term is positive.

Step 3. Find the required value

$S_{25}=5$ and $t_{25}=S_{25}-S_{24}=5-\sqrt{24}$, so

$$S_{25}-t_{25}=\sqrt{24}=2\sqrt6$$

Answer: option (A)

Arithmetic progression

A sequence is an arithmetic progression (AP) if every term minus the term before it is the same number $d$, the common difference:

$$a_{n+1}-a_n=d\ \text{ for every } n$$

The $n$th term

Starting from $a$, each step adds $d$, and reaching the $n$th term takes $n-1$ steps:

$$a_n=a+(n-1)d$$

So $a_n=dn+(a-d)$ is linear in $n$, and the points $(n,a_n)$ lie on a straight line of slope $d$. Conversely, any sequence with $a_n=An+B$ is an AP with $d=A$.

Terms of an AP plotted against n lie on one straight line, each step of 1 in n raising the term by d

In an AP of $m$ terms with last term $l$, the $n$th term from the end is the $(m-n+1)$th from the start:

$$a_{m-n+1}=l-(n-1)d$$

Sum of $n$ terms

Write the sum forwards and backwards and add. Each column adds to $a_1+a_n$, and there are $n$ columns:

$$2S_n=n(a_1+a_n)$$ $$S_n=\frac{n}{2}(a+l)=\frac{n}{2}\big[2a+(n-1)d\big]$$ Six bars of an AP with the same bars reversed on top, filling a rectangle of width n and height a1 plus an

Expanding, $S_n=\dfrac{d}{2}n^2+\left(a-\dfrac d2\right)n$, a quadratic in $n$ with no constant term. Conversely, if $S_n=An^2+Bn$, the sequence is an AP with $d=2A$. If $S_n=An^2+Bn+C$ with $C\ne0$, the terms from the second onward form an AP but $a_1$ does not fit it.

Properties of an AP

The last row works because $S_n=n\cdot a_{\frac{n+1}{2}}$ for odd $n$: the sum of an odd number of terms is the number of terms times the middle term.

Choosing the terms

When the sum of a few terms is given, choose them symmetrically so that $d$ cancels in the sum.

One arithmetic mean, $n$ arithmetic means

$A$ is the arithmetic mean of $a$ and $b$ when $a,A,b$ are in AP, so $A=\dfrac{a+b}{2}$. Inserting $n$ means $A_1,\dots,A_n$ makes an AP of $n+2$ terms with $b$ as the last term:

$$d=\frac{b-a}{n+1},\qquad A_k=a+kd$$ $$A_1+A_2+\dots+A_n=\frac{n(a+b)}{2}$$

Examples

Example 4 (JEE Main 2026, 28 Jan Shift 1). The common difference of the A.P. $a_1,a_2,\ldots,a_m$ is $13$ more than the common difference of the A.P. $b_1,b_2,\ldots,b_n$. If $b_{31}=-277$, $b_{43}=-385$ and $a_{78}=327$, then $a_1$ is equal to

▸ SHOW SOLUTION

Step 1. Difference of the second AP

$$b_{43}-b_{31}=12d_2=-108\Rightarrow d_2=-9$$

Step 2. Difference of the first AP

$$d_1=d_2+13=4$$

Step 3. First term

$$a_1=a_{78}-77d_1=327-308=19$$

Answer: option (3)

Example 5 (JEE Main 2026, 22 Jan Shift 1). If the sum of the first four terms of an A.P. is $6$ and the sum of its first six terms is $4$, then the sum of its first twelve terms is

▸ SHOW SOLUTION

Step 1. Two equations

$$S_4=2(2a+3d)=6\Rightarrow2a+3d=3$$$$S_6=3(2a+5d)=4\Rightarrow2a+5d=\frac43$$

Step 2. Solve

Subtracting, $2d=-\dfrac53$, so $d=-\dfrac56$ and $2a=3-3d=\dfrac{11}{2}$.

Step 3. Sum of twelve terms

$$S_{12}=6(2a+11d)$$$$=6\left(\frac{11}{2}-\frac{55}{6}\right)=-22$$

Answer: option (4)

Example 6 (JEE Main 2026, 24 Jan Shift 1). Consider an A.P. $a_1,a_2,\ldots,a_n$ with $a_1>0$. If $a_2-a_1=-\dfrac34$, $a_n=\dfrac14a_1$ and $\displaystyle\sum_{i=1}^{n}a_i=\frac{525}{2}$, then $\displaystyle\sum_{i=1}^{17}a_i$ is equal to

▸ SHOW SOLUTION

Step 1. Use the sum with first and last term

$$\frac{n}{2}\left(a_1+\frac{a_1}{4}\right)=\frac{525}{2}\Rightarrow na_1=420$$

Step 2. Use the last term

$$\frac{a_1}{4}=a_1-\frac34(n-1)\Rightarrow n-1=a_1$$

Step 3. Solve for $n$

$$n(n-1)=420\Rightarrow n=21,\ a_1=20$$

Step 4. Sum of seventeen terms

$$S_{17}=\frac{17}{2}\left(40-12\right)=238$$

Answer: option (3)

Example 7 (JEE Main 2026, 4 April Shift 1). The first term of an A.P. of $30$ non-negative terms is $\dfrac{10}{3}$. If the sum of this A.P. is the cube of its last term, then its common difference is

▸ SHOW SOLUTION

Step 1. Write the sum through the last term $L$

$$S_{30}=15\left(\frac{10}{3}+L\right)=15L+50$$

Step 2. Solve $15L+50=L^3$

$$L^3-15L-50=0$$$$(L-5)(L^2+5L+10)=0$$

The quadratic factor has discriminant $25-40<0$, so $L=5$.

Step 3. Common difference

$$\frac{10}{3}+29d=5\Rightarrow d=\frac{5}{87}$$

Answer: option (1)

Example 8 (JEE Main 2026, 24 Jan Shift 2). Let $\alpha_1,\alpha_2,\alpha_3,\alpha_4$ be an A.P. of four terms such that each term of the A.P. and its common difference $l$ are integers. If $\alpha_1+\alpha_2+\alpha_3+\alpha_4=48$ and $\alpha_1\alpha_2\alpha_3\alpha_4+l^4=361$, then the largest term of the A.P. is equal to

▸ SHOW SOLUTION

Step 1. Choose the four terms symmetrically

Take $a-3d,\ a-d,\ a+d,\ a+3d$, so $l=2d$. The sum is $4a=48$, so $a=12$.

Step 2. Write the product condition

$$(144-9d^2)(144-d^2)+16d^4=361$$$$25d^4-1440d^2+144^2=361$$

Step 3. Recognise a perfect square

$$\left(5d^2-144\right)^2=19^2$$

$5d^2=163$ gives no integer, so $5d^2=125$, $d=\pm5$ and $l=\pm10$.

Step 4. Largest term

The terms are $-3,7,17,27$ in some order, so the largest is $27$.

Answer: option (1)

Example 9 (JEE Main 2025, 22 Jan Shift 2). Suppose that the number of terms in an A.P. is $2k$, $k\in\mathbb{N}$. If the sum of all odd terms of the A.P. is $40$, the sum of all even terms is $55$, and the last term of the A.P. exceeds the first term by $27$, then $k$ is equal to

▸ SHOW SOLUTION

Step 1. Even terms minus odd terms

Pair $a_2$ with $a_1$, $a_4$ with $a_3$, and so on. Each pair differs by $d$ and there are $k$ pairs:

$$55-40=kd\Rightarrow kd=15$$

Step 2. Last term minus first term

$$(2k-1)d=27\Rightarrow2kd-d=27$$

Step 3. Solve

$30-d=27$, so $d=3$ and $k=5$.

Answer: option (1)

Example 10 (JEE Main 2025, 29 Jan Shift 2). Let $a_1,a_2,\dots,a_{2024}$ be an Arithmetic Progression such that $a_1+\left(a_5+a_{10}+a_{15}+\cdots+a_{2020}\right)+a_{2024}=2233$. Then $a_1+a_2+a_3+\cdots+a_{2024}$ is equal to ______.

▸ SHOW SOLUTION

Step 1. Count the bracket

$a_5,a_{10},\dots,a_{2020}$ has $\frac{2020}{5}=404$ terms.

Step 2. Pair the terms equidistant from the ends

$5+2020=10+2015=\dots=1+2024$, so each of the $202$ pairs in the bracket adds to $s=a_1+a_{2024}$:

$$s+202s=2233\Rightarrow s=11$$

Step 3. Required sum

$$S_{2024}=\frac{2024}{2}\,s=1012\cdot11=11132$$

Answer: $11132$

Example 11 (JEE Main 2025, 28 Jan Shift 2). The interior angles of a polygon with $n$ sides are in an A.P. with common difference $6°$. If the largest interior angle of the polygon is $219°$, then $n$ is equal to ______.

▸ SHOW SOLUTION

Step 1. Write the angles from the largest

The angles are $219,\ 213,\ \dots$ in degrees: an AP with first term $219$ and difference $-6$.

Step 2. Equate with the angle sum of a polygon

$$\frac{n}{2}\big[438-6(n-1)\big]=180(n-2)$$$$n^2-14n-120=0$$

Step 3. Solve

$(n-20)(n+6)=0$, so $n=20$. The smallest angle is $219-19\cdot6=105°$, which is positive.

Answer: $20$

Example 12 (JEE Main 2020, 7 Jan Shift 1). Five numbers are in A.P., whose sum is $25$ and product is $2520$. If one of these five numbers is $-\dfrac12$, then the greatest number amongst them is

▸ SHOW SOLUTION

Step 1. Choose the terms

Take $a-2d,\ a-d,\ a,\ a+d,\ a+2d$. The sum $5a=25$ gives $a=5$.

Step 2. Product condition

$$5(25-d^2)(25-4d^2)=2520$$

Step 3. Use the term $-\frac12$

$5-d=-\frac12$ gives $d=\frac{11}{2}$, and then $(25-d^2)(25-4d^2)=\left(-\frac{21}{4}\right)(-96)=504$, which matches $\frac{2520}{5}$. The choice $5-2d=-\frac12$ gives a negative product, so it fails.

Step 4. Greatest term

$$a+2d=5+11=16$$

Answer: option (4)

Example 13 (JEE Main 2026, 5 April Shift 1). Let the sum of the first $n$ terms of an A.P. be $3n^2+5n$. Then the sum of squares of the first $10$ terms of the A.P. is

▸ SHOW SOLUTION

Step 1. The $n$th term

$$T_n=S_n-S_{n-1}=6n+2$$

Step 2. Square and sum

$$\sum_{n=1}^{10}(6n+2)^2=36\sum n^2+24\sum n+40$$

Step 3. Substitute $\sum n^2=385$ and $\sum n=55$

$$36(385)+24(55)+40=15220$$

Answer: option (3)

Example 14 (JEE Main 2026, 2 April Shift 1). Let $A$ be the set of first $101$ terms of an A.P. whose first term is $1$ and common difference is $5$, and let $B$ be the set of first $71$ terms of an A.P. whose first term is $9$ and common difference is $7$. Then the number of elements in $A\cap B$ which are divisible by $3$ is

▸ SHOW SOLUTION

Step 1. Common terms form an AP

The first common term is $16$, and common terms repeat every $\operatorname{lcm}(5,7)=35$.

Two number lines marking the terms of A and of B, with the common terms 16, 51 and 86 joined by dashed lines

Step 2. Range

$A$ ends at $1+100\cdot5=501$ and $B$ at $9+70\cdot7=499$, so $16+35k\le499$ gives $k=0,1,\dots,13$.

Step 3. Divisible by $3$

$16+35k\equiv1+2k\pmod 3$, which is $0$ when $k\equiv1\pmod3$: $k=1,4,7,10,13$.

Answer: $5$ elements, option (2)

Example 15 (JEE Advanced 2018, Paper 1). Let $X$ be the set consisting of the first $2018$ terms of the arithmetic progression $1,6,11,\ldots$, and $Y$ be the set consisting of the first $2018$ terms of the arithmetic progression $9,16,23,\ldots$. Then, the number of elements in the set $X\cup Y$ is ______.

▸ SHOW SOLUTION

Step 1. Last terms

$X$ ends at $1+2017\cdot5=10086$ and $Y$ at $9+2017\cdot7=14128$.

Step 2. Common terms

They are $16+35m$, and must not exceed $10086$:

$$35m\le10070\Rightarrow m\le287$$

So there are $288$ common terms.

Step 3. Union

$$n(X\cup Y)=2018+2018-288=3748$$

Answer: $3748$

Example 16 (JEE Advanced 2015, Paper 2). Suppose that all the terms of an arithmetic progression (A.P.) are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is $6:11$ and the seventh term lies in between $130$ and $140$, then the common difference of this A.P. is ______.

▸ SHOW SOLUTION

Step 1. Use the middle-term form of the sums

$S_7=7a_4$ and $S_{11}=11a_6$, so

$$\frac{7(a+3d)}{11(a+5d)}=\frac{6}{11}\Rightarrow a=9d$$

Step 2. Seventh term

$$a_7=a+6d=15d$$

Step 3. Bounds

$130<15d<140$ and $d$ is a natural number, so $d=9$.

Answer: $9$

Example 17 (IIT JEE 2011, Paper 1). Let $a_1,a_2,a_3,\ldots,a_{100}$ be an arithmetic progression with $a_1=3$ and $S_p=\displaystyle\sum_{i=1}^{p}a_i$, $1\le p\le100$. For any integer $n$ with $1\le n\le20$, let $m=5n$. If $\dfrac{S_m}{S_n}$ does not depend on $n$, then $a_2$ is ______.

▸ SHOW SOLUTION

Step 1. Write the ratio

$$\frac{S_{5n}}{S_n}=\frac{5\big[(6-d)+5nd\big]}{(6-d)+nd}$$

Step 2. Make it free of $n$

A ratio $\frac{p+5nd}{p+nd}$ is constant in $n$ only when $p=0$ or $d=0$. Here $p=6-d$, so $d=6$ or $d=0$.

Step 3. Second term

$d=6$ gives $a_2=9$. The constant AP ($d=0$) gives $a_2=3$, where every ratio is $5$.

Answer: $9$ (the constant AP also allows $3$, so the answer is not unique)

Example 18 (JEE Advanced 2013, Paper 1). A pack contains $n$ cards numbered from $1$ to $n$. Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is $1224$. If the smaller of the numbers on the removed cards is $k$, then $k-20=$ ______.

▸ SHOW SOLUTION

Step 1. Set up

$$\frac{n(n+1)}{2}-(2k+1)=1224$$

Step 2. Bound $n$

Since $1\le k\le n-1$, the removed sum $2k+1$ lies between $3$ and $2n-1$, so $\frac{n(n+1)}{2}$ lies between $1227$ and $1223+2n$. $n=49$ gives $1225$, too small; $n=50$ gives $1275\le1323$; $n=51$ gives $1326>1325$, and larger $n$ only widen the gap. So $n=50$.

Step 3. Find $k$

$$2k+1=1275-1224=51\Rightarrow k=25$$

Answer: $k-20=5$

Example 19 (JEE Advanced 2022, Paper 1). Let $l_1,l_2,\ldots,l_{100}$ be consecutive terms of an arithmetic progression with common difference $d_1$, and let $w_1,w_2,\ldots,w_{100}$ be consecutive terms of another arithmetic progression with common difference $d_2$, where $d_1d_2=10$. For each $i=1,2,\ldots,100$, let $R_i$ be a rectangle with length $l_i$, width $w_i$ and area $A_i$. If $A_{51}-A_{50}=1000$, then the value of $A_{100}-A_{90}$ is ______.

▸ SHOW SOLUTION

Step 1. Difference of consecutive areas

$$A_{51}-A_{50}=l_{50}d_2+w_{50}d_1+d_1d_2$$$$l_{50}d_2+w_{50}d_1+10=1000$$

So $l_{50}d_2+w_{50}d_1=990$.

Step 2. Difference ten steps apart

$$A_{100}-A_{90}=10\left(l_{90}d_2+w_{90}d_1\right)$$$$\qquad+100d_1d_2$$

Step 3. Return to index $50$

$l_{90}=l_{50}+40d_1$ and $w_{90}=w_{50}+40d_2$, so

$$l_{90}d_2+w_{90}d_1=990+80d_1d_2=1790$$$$A_{100}-A_{90}=17900+1000=18900$$

Answer: $18900$

Example 20 (JEE Main 2026, 22 Jan Shift 2). Among the statements:

(S1): If $A(5,-1)$ and $B(-2,3)$ are two vertices of a triangle whose orthocentre is $(0,0)$, then its third vertex is $(-4,-7)$; and

(S2): If positive numbers $2a,b,c$ are three consecutive terms of an A.P., then the lines $ax+by+c=0$ are concurrent at $(2,-2)$.

▸ SHOW SOLUTION

Step 1. Check (S1)

For $C(h,k)$, $OA\perp BC$ and $OB\perp AC$:

$$5(h+2)-(k-3)=0$$$$-2(h-5)+3(k+1)=0$$

These give $k=5h+13$ and $2h-3k=13$, so $h=-4$, $k=-7$. (S1) is correct.

Step 2. Check (S2)

$2a,b,c$ in AP means $2b=2a+c$, that is $2a-2b+c=0$. This says exactly that $(2,-2)$ satisfies $ax+by+c=0$, for every such triple. (S2) is correct.

Answer: option (4)

Example 21 (JEE Main 2026, 4 April Shift 2). From a month of $31$ days, $3$ different dates are selected at random. If the probability that these dates are in an increasing A.P. is $\dfrac{a}{b}$, where $a,b\in\mathbb{N}$ and $\gcd(a,b)=1$, then $a+b$ is equal to ______.

▸ SHOW SOLUTION

Step 1. Count the AP triples

Three numbers $p<q<r$ are in AP when $q=\frac{p+r}{2}$, which needs $p$ and $r$ of the same parity. Each such pair fixes exactly one triple. Among $1,\dots,31$ there are $16$ odd and $15$ even numbers:

$$\binom{16}{2}+\binom{15}{2}=120+105=225$$

Step 2. Probability

$$\frac{225}{\binom{31}{3}}=\frac{225}{4495}=\frac{45}{899}$$

Answer: $a+b=944$

Example 22 (JEE Advanced 2026, Paper 2). Let $a,b,c$ be positive integers in arithmetic progression such that the equation $ax^2+bx+c=0$ has only integer solutions. Then which of the following statements is (are) TRUE?

▸ SHOW SOLUTION

Step 1. Write $b$ and $c$ through the roots $p,q$

$$b=-a(p+q),\qquad c=apq$$

Step 2. Use $2b=a+c$

$$-2(p+q)=1+pq$$$$(p+2)(q+2)=3$$

Step 3. Integer solutions

$\{p+2,q+2\}=\{1,3\}$ or $\{-1,-3\}$. The first gives $p+q=0$ and $b=0$, which is not allowed. So the roots are $-3$ and $-5$, with $b=8a$ and $c=15a$.

Step 4. Test the statements

(A)$c-b=7a$, true.

(B)$-3$ and $-5$ are odd, true.

(C)$c=15$ gives $a=1$, $b=8$, true.

(D)the roots are negative, so $3$ is not a root.

Answer: (A), (B) and (C)

Geometric progression

A sequence of non-zero numbers is a geometric progression (GP) if every term divided by the term before it is the same number $r$, the common ratio:

$$\frac{a_{n+1}}{a_n}=r\ \text{ for every } n$$

The $n$th term

Starting from $a$, each step multiplies by $r$:

$$a_n=ar^{\,n-1}$$

For $r>1$ and $a>0$ the terms grow faster and faster; for $0<r<1$ they shrink towards $0$; for $r<0$ the signs alternate.

Terms of a GP plotted against n: growing for r equal to 1.5 and decaying for r equal to 0.6

In a GP of $m$ terms with last term $l$, the $n$th term from the end is $l\left(\frac1r\right)^{n-1}$.

Sum of $n$ terms

Multiply $S_n=a+ar+\dots+ar^{n-1}$ by $r$ and subtract. All terms except the first and the last cancel:

$$S_n-rS_n=a-ar^{\,n}$$ $$S_n=\frac{a\left(r^{\,n}-1\right)}{r-1},\quad r\ne1$$

For $r=1$ every term is $a$ and $S_n=na$. In terms of the last term $l=ar^{n-1}$, $S_n=\dfrac{lr-a}{r-1}$.

Infinite GP

If $-1<r<1$, then $r^n\to0$ as $n\to\infty$, so $S_n$ approaches a fixed number:

$$S_\infty=\frac{a}{1-r},\quad \lvert r\rvert<1$$

For $\lvert r\rvert\ge1$ (and $a\ne0$) the terms do not shrink to $0$ and there is no finite sum.

Partial sums of a GP with ratio one half rising to a over one minus r, and with ratio minus one half swinging above and below it

Properties of a GP

The logarithm row links the two progressions: $\log a_n=\log a+(n-1)\log r$ is linear in $n$.

Choosing the terms

When a product is given, choose the terms so that $r$ cancels in the product.

Examples

Example 23 (JEE Main 2026, 21 Jan Shift 2). Let $\dfrac{a_2}{2},\dfrac{a_3}{2^2},\ldots,\dfrac{a_{10}}{2^9}$ be a G.P. of common ratio $\dfrac{1}{\sqrt2}$. If $a_1+a_2+\cdots+a_{10}=62$, then $a_1$ is equal to

▸ SHOW SOLUTION

Step 1. Ratio of the $a_k$

$$\frac{a_{k+1}/2^k}{a_k/2^{k-1}}=\frac{a_{k+1}}{2a_k}=\frac{1}{\sqrt2}\Rightarrow\frac{a_{k+1}}{a_k}=\sqrt2$$

So $a_1,\dots,a_{10}$ is a GP with ratio $\sqrt2$.

Step 2. Sum

$$\frac{a_1\left((\sqrt2)^{10}-1\right)}{\sqrt2-1}=\frac{31a_1}{\sqrt2-1}=62$$$$a_1=2(\sqrt2-1)$$

Answer: option (1)

Example 24 (JEE Main 2026, 21 Jan Shift 1). Let $a_1,a_2,a_3,\ldots$ be a G.P. of increasing positive terms such that $a_2a_3a_4=64$ and $a_1+a_3+a_5=\dfrac{813}{7}$. Then $a_3+a_5+a_7$ is equal to

▸ SHOW SOLUTION

Step 1. The product fixes the middle term

$$a_2a_3a_4=a_3^3=64\Rightarrow a_3=4$$

Step 2. The second condition

With $u=r^2$, $a_1+a_3+a_5=4\left(\frac1u+1+u\right)$:

$$28u^2-785u+28=0\Rightarrow u=28\ \text{or}\ \frac{1}{28}$$

The terms increase, so $r^2=28$.

Step 3. Required sum

$$a_3+a_5+a_7=4(1+28+784)=3252$$

Answer: option (2)

Example 25 (JEE Main 2026, 28 Jan Shift 1). In a G.P., if the product of the first three terms is $27$ and the set of all possible values for the sum of its first three terms is $\mathbb{R}-(a,b)$, then $a^2+b^2$ is equal to ______.

▸ SHOW SOLUTION

Step 1. Choose the terms

Take $\frac Ar,\ A,\ Ar$. The product is $A^3=27$, so $A=3$ and

$$S=3\left(1+r+\frac1r\right)$$

Step 2. Range of $r+\frac1r$

For $r>0$, $r+\frac1r\ge2$; for $r<0$, $r+\frac1r\le-2$.

Step 3. Range of $S$

$S\ge9$ or $S\le-3$, so $S\in\mathbb{R}-(-3,9)$.

Answer: $a^2+b^2=9+81=90$

Example 26 (JEE Main 2026, 6 April Shift 2). Consider the quadratic equation $(n^2-2n+2)x^2-3x+(n^2-2n+2)^2=0$, $n\in\mathbb{R}$. Let $\alpha$ be the minimum value of the product of its roots and $\beta$ be the maximum value of the sum of its roots. Then the sum of the first six terms of the G.P. whose first term is $\alpha$ and common ratio is $\dfrac{\alpha}{\beta}$ is

▸ SHOW SOLUTION

Step 1. Find $\alpha$ and $\beta$

Put $k=n^2-2n+2=(n-1)^2+1\ge1$. The product of roots is $k$, least value $\alpha=1$. The sum of roots is $\frac3k$, greatest value $\beta=3$.

Step 2. Sum of six terms

First term $1$, ratio $\frac13$:

$$S_6=\frac{1-3^{-6}}{1-\frac13}=\frac{3}{2}\cdot\frac{728}{729}=\frac{364}{243}$$

Answer: option (3)

Example 27 (JEE Main 2026, 6 April Shift 1). For the functions $f(\theta)=\alpha\tan^2\theta+\beta\cot^2\theta$ and $g(\theta)=\alpha\sin^2\theta+\beta\cos^2\theta$, $\alpha>\beta>0$, let $\displaystyle\min_{0<\theta<\pi/2}f(\theta)=\max_{0<\theta<\pi}g(\theta)$. If the first term of a G.P. is $\left(\dfrac{\alpha}{2\beta}\right)$, its common ratio is $\left(\dfrac{2\beta}{\alpha}\right)$, and the sum of its first $10$ terms is $\dfrac{m}{n}$ with $\gcd(m,n)=1$, then $m+n$ is equal to ______.

▸ SHOW SOLUTION

Step 1. Minimum of $f$

The two terms have product $\alpha\beta$, so by AM and GM, $f\ge2\sqrt{\alpha\beta}$, with equality when $\tan^4\theta=\frac\beta\alpha$.

Step 2. Maximum of $g$

$g=(\alpha-\beta)\sin^2\theta+\beta$, largest value $\alpha$ at $\theta=\frac\pi2$.

Step 3. Equate

$$2\sqrt{\alpha\beta}=\alpha\Rightarrow\alpha=4\beta$$

So the GP has first term $2$ and ratio $\frac12$.

Step 4. Sum of ten terms

$$S_{10}=\frac{2\left(1-2^{-10}\right)}{1/2}=\frac{1023}{256}$$

Answer: $m+n=1279$

Example 28 (JEE Main 2026, 22 Jan Shift 2). Let $f$ and $g$ be functions satisfying $f(x+y)=f(x)f(y)$ and $g(x+y)=g(xy)$, with $f(1)=7$ and $g(1)=1$, for all $x,y\in\mathbb{N}$. If $\displaystyle\sum_{x=1}^{n}\frac{f(x)}{g(x)}=19607$, then $n$ is equal to

▸ SHOW SOLUTION

Step 1. Find $f$ and $g$

$y=1$ gives $f(x+1)=7f(x)$, so $f(x)=7^x$. Also $g(x+1)=g(x)$, so $g(x)=g(1)=1$.

Step 2. The sum is a GP

$$\sum_{x=1}^{n}7^x=\frac{7\left(7^n-1\right)}{6}=19607$$

Step 3. Solve

$7^n-1=16806$, so $7^n=16807=7^5$.

Answer: $n=5$, option (2)

Example 29 (JEE Main 2026, 28 Jan Shift 2). $\dfrac{6}{3^{26}}+\dfrac{10\cdot1}{3^{25}}+\dfrac{10\cdot2}{3^{24}}+\dfrac{10\cdot2^2}{3^{23}}+\cdots+\dfrac{10\cdot2^{24}}{3}$ is equal to

▸ SHOW SOLUTION

Step 1. The tail is a GP

$$\frac{10\cdot2^j}{3^{25-j}}=\frac{10}{3^{25}}\cdot6^j,\quad j=0,1,\dots,24$$

Step 2. Sum the tail

$$\frac{10}{3^{25}}\cdot\frac{6^{25}-1}{5}=\frac{2\cdot6^{25}-2}{3^{25}}$$

Step 3. Add the first term

$\frac{6}{3^{26}}=\frac{2}{3^{25}}$, which cancels the $-2$:

$$S=\frac{2\cdot6^{25}}{3^{25}}=2\cdot2^{25}=2^{26}$$

Answer: option (2)

Example 30 (JEE Main 2026, 24 Jan Shift 1). Let $729,81,9,1,\ldots$ be a sequence and $P_n$ denote the product of the first $n$ terms of this sequence. If $2\displaystyle\sum_{n=1}^{40}(P_n)^{1/n}=\frac{3^\alpha-1}{3^\beta}$ and $\gcd(\alpha,\beta)=1$, then $\alpha+\beta$ is equal to

▸ SHOW SOLUTION

Step 1. Exponents form an AP

$a_n=3^{8-2n}$, so $P_n=3^{6+4+\dots+(8-2n)}$ and the exponent is

$$\frac n2\big[6+(8-2n)\big]=n(7-n)$$

Step 2. The $n$th root

$$(P_n)^{1/n}=3^{7-n}$$

Step 3. Sum the GP

$$\sum_{n=1}^{40}3^{7-n}=3^6\cdot\frac{1-3^{-40}}{2/3}=\frac{3^{40}-1}{2\cdot3^{33}}$$

So $2\sum=\frac{3^{40}-1}{3^{33}}$, $\alpha=40$, $\beta=33$.

Answer: $\alpha+\beta=73$, option (1)

Example 31 (JEE Main 2026, 5 April Shift 2). Let $\alpha,\beta$ be the roots of $x^2-x+p=0$ and $\gamma,\delta$ be the roots of $x^2-4x+q=0$; $p,q\in\mathbb{Z}$. If $\alpha,\beta,\gamma,\delta$ are in G.P., then $\lvert p+q\rvert$ equals

▸ SHOW SOLUTION

Step 1. Write the four roots

Let them be $a,ar,ar^2,ar^3$. Then $a(1+r)=1$ and $ar^2(1+r)=4$, so $r^2=4$.

Step 2. Test $r=2$

$a=\frac13$ and $p=a^2r=\frac29$, not an integer. Reject.

Step 3. Take $r=-2$

$a(1-2)=1$ gives $a=-1$:

$$p=a^2r=-2,\qquad q=a^2r^5=-32$$

Answer: $\lvert p+q\rvert=34$, option (3)

Example 32 (JEE Main 2021, 25 Feb Shift 1). Let $A_1,A_2,A_3,\ldots$ be squares such that for each $n\ge1$, the length of the side of $A_n$ equals the length of the diagonal of $A_{n+1}$. If the length of the side of $A_1$ is $12$ cm, then the smallest value of $n$ for which the area of $A_n$ is less than one, is ______.

▸ SHOW SOLUTION

Step 1. Ratio of sides

A square of side $s$ has diagonal $s\sqrt2$. Here side of $A_n=\sqrt2\times$ side of $A_{n+1}$, so the sides form a GP with ratio $\frac{1}{\sqrt2}$ and the areas a GP with ratio $\frac12$.

Four nested squares, each one turned through half a right angle inside the previous one, beside bars of their areas halving from 144

Step 2. Area of $A_n$

$$\text{area}(A_n)=\frac{144}{2^{\,n-1}}$$

Step 3. Least $n$

$2^{n-1}>144$ needs $n-1\ge8$, since $2^7=128$ and $2^8=256$.

Answer: $n=9$

Example 33 (JEE Advanced 2014, Paper 1). Let $a,b,c$ be positive integers such that $\dfrac ba$ is an integer. If $a,b,c$ are in geometric progression and the arithmetic mean of $a,b,c$ is $b+2$, then the value of $\dfrac{a^2+a-14}{a+1}$ is ______.

▸ SHOW SOLUTION

Step 1. Write the terms

$b=ar$ and $c=ar^2$, with $r$ a positive integer.

Step 2. Use the mean

$$a+ar+ar^2=3ar+6\Rightarrow a(r-1)^2=6$$

Step 3. Integer solution

$(r-1)^2$ must be a square that divides $6$, so $(r-1)^2=1$. Then $r=2$ (since $r=0$ makes $b=0$) and $a=6$.

Step 4. Evaluate

$$\frac{36+6-14}{7}=4$$

Answer: $4$

Example 34 (JEE Advanced 2025, Paper 1). Let $\mathbb{R}$ denote the set of all real numbers. Let $f:\mathbb{R}\to\mathbb{R}$ be a function such that $f(x)>0$ for all $x\in\mathbb{R}$, and $f(x+y)=f(x)f(y)$ for all $x,y\in\mathbb{R}$. Let the real numbers $a_1,a_2,\ldots,a_{50}$ be in an arithmetic progression. If $f(a_{31})=64f(a_{25})$, and $\displaystyle\sum_{i=1}^{50}f(a_i)=3\left(2^{25}+1\right)$, then the value of $\displaystyle\sum_{i=6}^{30}f(a_i)$ is ______.

▸ SHOW SOLUTION

Step 1. $f$ turns an AP into a GP

$f(a_{i+1})=f(a_i+d)=f(a_i)f(d)$, so $f(a_1),f(a_2),\dots$ is a GP with ratio $q=f(d)>0$.

Step 2. Find the ratio

$$f(a_{31})=f(a_{25})\,q^6=64f(a_{25})\Rightarrow q=2$$

Step 3. Use the total

$$f(a_1)\left(2^{50}-1\right)=3\left(2^{25}+1\right)$$

$2^{50}-1=(2^{25}-1)(2^{25}+1)$, so $f(a_1)=\frac{3}{2^{25}-1}$.

Step 4. Required sum

$$\sum_{i=6}^{30}f(a_i)=f(a_1)\cdot2^5\left(2^{25}-1\right)=96$$

Answer: $96$

Example 35 (JEE Advanced 2016, Paper 2). Let $b_i>1$ for $i=1,2,\ldots,101$. Suppose $\log_eb_1,\log_eb_2,\ldots,\log_eb_{101}$ are in Arithmetic Progression (A.P.) with the common difference $\log_e2$. Suppose $a_1,a_2,\ldots,a_{101}$ are in A.P. such that $a_1=b_1$ and $a_{51}=b_{51}$. If $t=b_1+b_2+\cdots+b_{51}$ and $s=a_1+a_2+\cdots+a_{51}$, then

▸ SHOW SOLUTION

Step 1. The $b_i$ form a GP

$\log b_i$ is an AP with difference $\log2$, so $b_i=b_1\,2^{\,i-1}$.

Step 2. Compare $s$ and $t$

$$t=b_1\left(2^{51}-1\right)$$$$s=\frac{51}{2}\left(b_1+b_1\,2^{50}\right)>\frac{51}{2}\,b_1\,2^{50}>t$$

Step 3. Compare the $101$st terms

$a_{101}=2a_{51}-a_1=b_1\left(2^{51}-1\right)$, while $b_{101}=b_1\,2^{100}$, which is far larger.

Answer: option (B)

Examples on the infinite GP

Example 36 (JEE Main 2026, 4 April Shift 2). Let $\alpha=\dfrac14+\dfrac18+\dfrac1{16}+\cdots\infty$ and $\beta=\dfrac13+\dfrac19+\dfrac1{27}+\cdots\infty$. Then the value of $(0.2)^{\log_{\sqrt5}(\alpha)}+(0.04)^{\log_5(\beta)}$ is

▸ SHOW SOLUTION

Step 1. The two sums

$$\alpha=\frac{1/4}{1-1/2}=\frac12,\qquad\beta=\frac{1/3}{1-1/3}=\frac12$$

Step 2. First term

$0.2=(\sqrt5)^{-2}$, so

$$(0.2)^{\log_{\sqrt5}\alpha}=\left(\sqrt5^{\,\log_{\sqrt5}\alpha}\right)^{-2}=\alpha^{-2}=4$$

Step 3. Second term

$0.04=5^{-2}$, so $(0.04)^{\log_5\beta}=\beta^{-2}=4$.

Answer: $4+4=8$, option (3)

Example 37 (JEE Main 2019, 11 Jan Shift 1). The sum of an infinite geometric series with positive terms is $3$ and the sum of the cubes of its terms is $\dfrac{27}{19}$. Then the common ratio of this series is

▸ SHOW SOLUTION

Step 1. Two equations

The cubes form a GP with ratio $r^3$:

$$\frac{a}{1-r}=3,\qquad\frac{a^3}{1-r^3}=\frac{27}{19}$$

Step 2. Eliminate $a$

Put $a=3(1-r)$ and cancel $1-r$:

$$\frac{(1-r)^2}{1+r+r^2}=\frac{1}{19}\Rightarrow6r^2-13r+6=0$$

Step 3. Choose the root

$r=\frac23$ or $\frac32$, and an infinite sum needs $r<1$.

Answer: option (2)

Example 38 (JEE Main 2026, 24 Jan Shift 2). The sum

$$\left(\frac13+\frac47\right)+\left(\frac{1}{3^2}+\frac13\cdot\frac47+\frac{4^2}{7^2}\right)+\cdots$$

where the $k$th bracket holds every product $\left(\frac13\right)^{i}\left(\frac47\right)^{j}$ with $i+j=k$, up to infinite terms, is equal to

▸ SHOW SOLUTION

Step 1. Each bracket is a finite GP

With $a=\frac47$ and $b=\frac13$, the $k$th bracket is

$$\sum_{i+j=k}a^jb^i=\frac{a^{k+1}-b^{k+1}}{a-b}$$

Step 2. Add over $k\ge1$

$$S=\frac{1}{a-b}\left(\frac{a^2}{1-a}-\frac{b^2}{1-b}\right)$$

Step 3. Substitute

$\frac{a^2}{1-a}=\frac{16}{21}$, $\frac{b^2}{1-b}=\frac16$ and $a-b=\frac{5}{21}$:

$$S=\frac{21}{5}\cdot\frac{25}{42}=\frac52$$

Answer: option (1)

AP and GP together

Many questions name two progressions and link them. The working rule is the same every time: give each progression its own unknowns ($a,d$ for the AP; $b,r$ for the GP), turn every condition into an equation, and use the three-term tests $2b=a+c$ and $b^2=ac$.

Examples

Example 39 (JEE Main 2026, 2 April Shift 2). Let $a_1,a_2,a_3,\ldots$ be an A.P. and $g_1=a_1,g_2,g_3,\ldots$ be an increasing G.P. If $a_1=a_2+g_2=1$ and $a_3+g_3=4$, then $a_{10}+g_5$ is equal to

▸ SHOW SOLUTION

Step 1. Unknowns

$a_1=g_1=1$, $a_n=1+(n-1)d$, $g_n=r^{n-1}$.

Step 2. Two equations

$$(1+d)+r=1\Rightarrow d=-r$$$$(1+2d)+r^2=4\Rightarrow r^2-2r-3=0$$

Step 3. Choose $r$

$r=3$ or $r=-1$; the GP increases, so $r=3$ and $d=-3$.

Step 4. Required value

$$a_{10}+g_5=(1-27)+81=55$$

Answer: option (D)

Example 40 (JEE Main 2026, 6 April Shift 1). The sum of the first ten terms of an A.P. is $160$ and the sum of the first two terms of a G.P. is $8$. If the first term of the A.P. is equal to the common ratio of the G.P. and the first term of the G.P. is equal to the common difference of the A.P., then the sum of all possible values of the first term of the G.P. is

▸ SHOW SOLUTION

Step 1. Name the unknowns

AP: first term $a$, difference $d$. GP: first term $d$, ratio $a$.

$$5(2a+9d)=160\Rightarrow2a+9d=32$$$$d+da=8$$

Step 2. Eliminate $a$

$a=\frac{32-9d}{2}$, so

$$d\cdot\frac{34-9d}{2}=8\Rightarrow9d^2-34d+16=0$$

Step 3. Sum of roots

Both roots are real (discriminant $580>0$), and their sum is $\frac{34}{9}$.

Answer: option (1)

Example 41 (JEE Main 2026, 22 Jan Shift 2). Suppose $a,b,c$ are in A.P. and $a^2,2b^2,c^2$ are in G.P. If $a<b<c$ and $a+b+c=1$, then $9(a^2+b^2+c^2)$ is equal to ______.

▸ SHOW SOLUTION

Step 1. Choose the AP

$a=b-d$, $c=b+d$, $d>0$. The sum gives $b=\frac13$.

Step 2. GP condition

$$4b^4=a^2c^2\Rightarrow\left(b^2-d^2\right)^2=\frac{4}{81}$$

$b^2-d^2=\frac29$ gives $d^2<0$, so $b^2-d^2=-\frac29$ and $d^2=\frac13$.

Step 3. Sum of squares

$$a^2+b^2+c^2=3b^2+2d^2=\frac13+\frac23=1$$

Answer: $9$

Example 42 (JEE Main 2025, 7 April Shift 1). Let $x_1,x_2,x_3,x_4$ be in a geometric progression. If $2,7,9,5$ are subtracted respectively from $x_1,x_2,x_3,x_4$, then the resulting numbers are in an arithmetic progression. Then the value of $\dfrac{1}{24}\left(x_1x_2x_3x_4\right)$ is

▸ SHOW SOLUTION

Step 1. Equal differences

With $x_k=ar^{k-1}$, the numbers $a-2,\ ar-7,\ ar^2-9,\ ar^3-5$ are in AP:

$$a(r-1)-5=ar(r-1)-2$$$$ar(r-1)-2=ar^2(r-1)+4$$

Step 2. Subtract in pairs

$$a(r-1)^2=-3,\qquad ar(r-1)^2=-6$$

Dividing, $r=2$, and then $a=-3$.

Step 3. Product

The GP is $-3,-6,-12,-24$, with product $5184$.

Answer: $\frac{5184}{24}=216$, option (4)

Example 43 (JEE Main 2019, 10 Apr Shift 2). Let $a,b$ and $c$ be in G.P. with common ratio $r$, where $a\ne0$ and $0<r\le\dfrac12$. If $3a,7b$ and $15c$ are the first three terms of an A.P., then the $4$th term of this A.P. is

▸ SHOW SOLUTION

Step 1. AP condition

$$14ar=3a+15ar^2$$$$15r^2-14r+3=0$$

Step 2. Choose $r$

$r=\frac13$ or $\frac35$; only $\frac13\le\frac12$.

Step 3. The AP

$3a,\ \frac73a,\ \frac53a$ with difference $-\frac23a$, so the fourth term is $\frac53a-\frac23a=a$.

Answer: option (1)

Example 44 (JEE Advanced 2020, Paper 1). Let $a_1,a_2,a_3,\ldots$ be a sequence of positive integers in arithmetic progression with common difference $2$. Also, let $b_1,b_2,b_3,\ldots$ be a sequence of positive integers in geometric progression with common ratio $2$. If $a_1=b_1=c$, then the number of all possible values of $c$, for which the equality $2\left(a_1+a_2+\cdots+a_n\right)=b_1+b_2+\cdots+b_n$ holds for some positive integer $n$, is ______.

▸ SHOW SOLUTION

Step 1. Both sums

$$2\left[nc+n(n-1)\right]=c\left(2^n-1\right)$$$$c=\frac{2n(n-1)}{2^n-1-2n}$$

Step 2. Try small $n$

$n=1$ gives $c=0$ and $n=2$ gives a negative value. $n=3$ gives $c=\frac{12}{1}=12$. For $n=4,5,6$ the values are $\frac{24}{7},\frac{40}{21},\frac{60}{51}$, not integers.

Step 3. Large $n$

For $n\ge7$, $2^n-1-2n>2n(n-1)$, so $0<c<1$ and $c$ is not a positive integer.

Answer: $1$

Harmonic progression

A sequence of non-zero numbers is a harmonic progression (HP) if the reciprocals of its terms form an AP:

$$a_1,a_2,\dots\ \text{HP}\iff\frac{1}{a_1},\frac{1}{a_2},\dots\ \text{AP}$$

For example $1,\frac13,\frac15,\frac17,\dots$ is an HP because $1,3,5,7,\dots$ is an AP.

The $n$th term

There is no formula for the sum of an HP. Every HP question is solved by turning to the AP of reciprocals, doing the work there, and turning back. If the reciprocal AP has first term $\frac1a$ and difference $D$,

$$a_n=\frac{1}{\frac1a+(n-1)D}$$

Three terms in HP

$$a,b,c\ \text{in HP}\iff\frac2b=\frac1a+\frac1c$$ $$b=\frac{2ac}{a+c}$$

Two equivalent forms are often quicker:

$$\frac{a}{c}=\frac{a-b}{b-c}$$ $$\frac{b+a}{b-a}+\frac{b+c}{b-c}=2$$

Inserting harmonic means

To insert $H_1,\dots,H_n$ between $a$ and $b$, insert $n$ arithmetic means between $\frac1a$ and $\frac1b$:

$$D=\frac{a-b}{(n+1)ab},\qquad\frac{1}{H_k}=\frac1a+kD$$

Examples

Example 45. The fourth term of a harmonic progression is $\dfrac35$ and its ninth term is $\dfrac{3}{10}$. Then its fourteenth term is

▸ SHOW SOLUTION

Step 1. Pass to the reciprocal AP

The AP has fourth term $\frac53$ and ninth term $\frac{10}{3}$:

$$5D=\frac{10}{3}-\frac53\Rightarrow D=\frac13$$

Step 2. Fourteenth term of the AP

$$\frac{10}{3}+5\cdot\frac13=5$$

Step 3. Return to the HP

The fourteenth term of the HP is $\frac15$.

Answer: option (A)

Example 46 (IITianForum). Let $2,b,c$ be in A.P., $b,c,d$ be in G.P. and $c,d,e$ be in H.P. If $e=18$ and $c>0$, then $b+d$ is equal to

▸ SHOW SOLUTION

Step 1. Express $d$ and $e$ through $a=2$ and $c$

AP: $b=\frac{a+c}{2}$. GP: $d=\frac{c^2}{b}=\frac{2c^2}{a+c}$.

Step 2. HP condition

$$\frac2d=\frac1c+\frac1e\Rightarrow e=\frac{cd}{2c-d}$$

Substituting $d$:

$$e=\frac{2c^3/(a+c)}{2ac/(a+c)}=\frac{c^2}{a}$$

So $a,c,e$ are in GP, a result worth remembering.

Step 3. Values

$c^2=2\cdot18=36$ and $c>0$, so $c=6$, $b=4$, $d=\frac{36}{4}=9$. Check: $\frac2d=\frac29=\frac16+\frac1{18}$.

Answer: $b+d=13$, option (B)

Arithmetic, geometric and harmonic means

For two positive numbers $a$ and $b$, the three means are

$$A=\frac{a+b}{2},\quad G=\sqrt{ab},\quad H=\frac{2ab}{a+b}$$

$A$ is the middle term of an AP $a,A,b$; $G$ of a GP $a,G,b$; $H$ of an HP $a,H,b$.

On a number line from a equal to 2 to b equal to 8, the harmonic mean 3.2, the geometric mean 4 and the arithmetic mean 5

Results on the three means

The order $A\ge G$ comes from one square:

$$A-G=\frac{\left(\sqrt a-\sqrt b\right)^2}{2}\ge0$$

and $G\ge H$ follows from $G^2=AH$ with $A\ge G$.

Three arithmetic means between 1 and 16 equally spaced, and three geometric means 2, 4 and 8 crowding towards the smaller end

The inserted arithmetic means divide the gap from $a$ to $b$ into equal parts; the geometric means divide it in equal ratios, so they crowd towards the smaller number.

Examples

Example 47 (JEE Main 2023, 11 Apr Shift 1). Let $x_1,x_2,\ldots,x_{100}$ be in an arithmetic progression, with $x_1=2$ and their mean equal to $200$. If $y_i=i(x_i-i)$, $1\le i\le100$, then the mean of $y_1,y_2,\ldots,y_{100}$ is

▸ SHOW SOLUTION

Step 1. Common difference

The mean of an AP is the mean of its first and last terms:

$$\frac{2+x_{100}}{2}=200\Rightarrow x_{100}=398,\ d=4$$

Step 2. Write $y_i$

$x_i=4i-2$, so

$$y_i=i(3i-2)=3i^2-2i$$

Step 3. Mean

$$\frac{3(338350)-2(5050)}{100}=10049.5$$

Answer: option (3)

Example 48 (IITianForum). Let $A_i$, $G_i$ and $H_i$ $(i=1,2,\ldots,5)$ be the five arithmetic means, five geometric means and five harmonic means inserted between $3$ and $48$. Then $A_2H_4+\left(G_1G_2G_3G_4G_5\right)^{1/5}$ is equal to

▸ SHOW SOLUTION

Step 1. Arithmetic mean $A_2$

$d=\frac{48-3}{6}=\frac{15}{2}$, so $A_2=3+15=18$.

Step 2. Harmonic mean $H_4$

$$D=\frac{1}{6}\left(\frac1{48}-\frac13\right)=-\frac{5}{96}$$$$\frac{1}{H_4}=\frac13-\frac{20}{96}=\frac18$$

So $H_4=8$ and $A_2H_4=144=3\cdot48$, as the table says.

Step 3. Geometric means

The product of the five means is $\left(\sqrt{144}\right)^5$, so its fifth root is $12$.

Answer: $144+12=156$, option (C)

Arithmetico-geometric progression

Multiplying an AP term by term with a GP gives an arithmetico-geometric progression (AGP):

$$a,\ (a+d)r,\ (a+2d)r^2,\ \dots$$ $$T_n=\big[a+(n-1)d\big]r^{\,n-1}$$ Bars of the terms n times 0.6 to the power n minus 1, rising for two steps and then falling towards zero

Summing an AGP: multiply by $r$ and subtract

Write $S$, write $rS$ shifted by one place, and subtract. The AP parts differ by $d$ in every column, so what remains is a GP:

$$S-rS=a+\left(dr+dr^2+\dots+dr^{\,n-1}\right)$$ $$\qquad-\big[a+(n-1)d\big]r^{\,n}$$

For an infinite AGP with $\lvert r\rvert<1$, the last term goes to $0$:

$$S_\infty=\frac{a}{1-r}+\frac{dr}{(1-r)^2}$$

Two special cases are worth knowing by heart:

$$\sum_{k=1}^{\infty}kx^{k-1}=\frac{1}{(1-x)^2},\quad\lvert x\rvert<1$$ $$\sum_{k=1}^{n}k\,2^k=(n-1)2^{\,n+1}+2$$

Examples

Example 49 (JEE Main 2025, 24 Jan Shift 2). If $7=5+\dfrac17(5+\alpha)+\dfrac{1}{7^2}(5+2\alpha)+\dfrac{1}{7^3}(5+3\alpha)+\cdots\infty$, then the value of $\alpha$ is

▸ SHOW SOLUTION

Step 1. Split the series

$$\sum_{k\ge1}\frac{5+k\alpha}{7^k}=5\sum_{k\ge1}\frac{1}{7^k}+\alpha\sum_{k\ge1}\frac{k}{7^k}$$

Step 2. The two sums

$$\sum_{k\ge1}\frac{1}{7^k}=\frac16,\qquad\sum_{k\ge1}\frac{k}{7^k}=\frac{1/7}{(6/7)^2}=\frac{7}{36}$$

Step 3. Solve

$$7=5+\frac56+\frac{7\alpha}{36}\Rightarrow\frac{7\alpha}{36}=\frac76$$

Answer: $\alpha=6$, option (3)

Example 50 (IITianForum). The value of the infinite product $2^{1/2}\cdot4^{1/4}\cdot8^{1/8}\cdot16^{1/16}\cdots$ is

▸ SHOW SOLUTION

Step 1. Write every factor as a power of $2$

$$\left(2^k\right)^{1/2^k}=2^{k/2^k}$$

Step 2. Add the exponents

$$S=\frac12+\frac24+\frac38+\dots$$

This is an AGP. $\frac12S=\frac14+\frac28+\dots$, and subtracting,

$$\frac12S=\frac12+\frac14+\frac18+\dots=1$$

Step 3. Product

$S=2$, so the product is $2^2=4$.

Answer: option (C)

Example 51. The sum $1\cdot2+2\cdot2^2+3\cdot2^3+\cdots+10\cdot2^{10}$ is equal to

▸ SHOW SOLUTION

Step 1. Multiply by $2$ and subtract

$$S-2S=2+2^2+\dots+2^{10}-10\cdot2^{11}$$

Step 2. Sum the GP

$$-S=\left(2^{11}-2\right)-10\cdot2^{11}=-9\cdot2^{11}-2$$

Step 3. Value

$$S=9\cdot2048+2=18434$$

Answer: option (B)

Sums of $n$, $n^2$ and $n^3$

$$\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$$ $$\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}$$ $$\sum_{k=1}^{n}k^3=\left[\frac{n(n+1)}{2}\right]^2$$

Why the formulas hold

The first is the AP sum. For the second, add $(k+1)^3-k^3=3k^2+3k+1$ from $k=1$ to $n$; the left side collapses to $(n+1)^3-1$, and solving for $\sum k^2$ gives the formula. The third has a picture: a square of side $1+2+\dots+n$ splits into L-shaped strips, and the strip of width $k$ has area $k^3$.

A square of side 1 plus 2 plus 3 plus 4 split into L-shaped strips of widths 1 to 4, beside k copies of a k by k square for each k

Summing any polynomial term

If $T_k$ is a polynomial in $k$, expand it and use the three formulas term by term:

$$\sum_{k=1}^{n}\left(ak^3+bk^2+ck+e\right)$$ $$=a\sum k^3+b\sum k^2+c\sum k+ne$$

Examples

Example 52 (JEE Main 2026, 2 April Shift 2). The sum $\dfrac{1^3}{1}+\dfrac{1^3+2^3}{1+3}+\dfrac{1^3+2^3+3^3}{1+3+5}+\cdots$ up to $8$ terms, is

▸ SHOW SOLUTION

Step 1. General term

$$T_k=\frac{\left[\frac{k(k+1)}{2}\right]^2}{k^2}=\frac{(k+1)^2}{4}$$

Step 2. Sum

$$\sum_{k=1}^{8}\frac{(k+1)^2}{4}=\frac{2^2+3^2+\dots+9^2}{4}$$$$=\frac{285-1}{4}=71$$

Answer: $71$, option (B)

Example 53 (JEE Main 2026, 6 April Shift 2). The sum $1+\dfrac12\left(1^2+2^2\right)+\dfrac13\left(1^2+2^2+3^2\right)+\cdots$ up to $10$ terms is equal to

▸ SHOW SOLUTION

Step 1. General term

$$T_k=\frac1k\cdot\frac{k(k+1)(2k+1)}{6}$$$$T_k=\frac{2k^2+3k+1}{6}$$

Step 2. Sum

$$\sum_{k=1}^{10}T_k=\frac{2(385)+3(55)+10}{6}=\frac{945}{6}$$

Answer: $\frac{315}{2}$, option (3)

Example 54 (JEE Main 2026, 6 April Shift 1). The value of $1^3-2^3+3^3-\cdots+15^3$ is

▸ SHOW SOLUTION

Step 1. Take all cubes, then remove the even ones twice

$$\sum_{k=1}^{15}k^3-2\sum_{j=1}^{7}(2j)^3$$

Step 2. Evaluate

$$14400-16\cdot784=14400-12544$$

Answer: $1856$, option (2)

Example 55 (JEE Main 2019, 12 Jan Shift 2). If the sum of the first $15$ terms of the series $\left(\dfrac34\right)^3+\left(1\dfrac12\right)^3+\left(2\dfrac14\right)^3+3^3+\left(3\dfrac34\right)^3+\cdots$ is equal to $225K$, then $K$ is equal to

▸ SHOW SOLUTION

Step 1. General term

The bases are $\frac34,\frac64,\frac94,\dots$, so $T_k=\left(\frac{3k}{4}\right)^3$.

Step 2. Sum

$$\frac{27}{64}\left(\frac{15\cdot16}{2}\right)^2=\frac{27}{64}\cdot14400=6075$$

Answer: $K=\frac{6075}{225}=27$, option (2)

Example 56 (JEE Main 2026, 2 April Shift 1). Let $\alpha,\ \alpha+2$, $\alpha\in\mathbb{Z}$, be the roots of the quadratic equation $x(x+2)+(x+1)(x+3)+(x+2)(x+4)+\cdots+(x+n-1)(x+n+1)=4n$ for some $n\in\mathbb{N}$. Then $(n+\alpha)$ is equal to

▸ SHOW SOLUTION

Step 1. Sum the left side

$(x+k-1)(x+k+1)=(x+k)^2-1$. Adding for $k=1$ to $n$ and dividing by $n$:

$$x^2+(n+1)x+\frac{(n+1)(2n+1)}{6}=5$$

Step 2. Roots differ by $2$

For a monic quadratic, (difference of roots)$^2=B^2-4C$:

$$(n+1)^2-\frac{4(n+1)(2n+1)}{6}+20=4$$

This simplifies to $n^2=49$, so $n=7$.

Step 3. Roots

$x^2+8x+15=0$ gives $x=-5,-3$, so $\alpha=-5$.

Answer: $n+\alpha=2$, option (3)

Example 57 (JEE Main 2026, 21 Jan Shift 2). The positive integer $n$, for which the solutions of the equation $x(x+2)+(x+2)(x+4)+\cdots+(x+2n-2)(x+2n)=\dfrac{8n}{3}$ are two consecutive even integers, is

▸ SHOW SOLUTION

Step 1. Sum the left side

$(x+2k-2)(x+2k)=x^2+(4k-2)x+4k(k-1)$, and

$$\sum_{k=1}^{n}4k(k-1)=\frac{4(n-1)n(n+1)}{3}$$

Step 2. The quadratic

Dividing by $n$:

$$x^2+2nx+\frac{4n^2-12}{3}=0$$

Step 3. Roots differ by $2$

$$4n^2-\frac{4(4n^2-12)}{3}=4\Rightarrow n^2=9$$

So $n=3$, and the roots of $x^2+6x+8=0$ are $-4$ and $-2$.

Answer: option (1)

Example 58 (JEE Main 2026, 8 April Shift 2). Let $\alpha=3+4+8+9+13+14+\cdots$ upto $40$ terms. If $(\tan\beta)^{\alpha/1020}$ is a root of $x^2+x-2=0$, $\beta\in\left(0,\dfrac\pi2\right)$, then $\sin^2\beta+3\cos^2\beta$ is equal to

▸ SHOW SOLUTION

Step 1. Group in pairs

The pairs are $(3,4),(8,9),(13,14),\dots$, with sums $7,17,27,\dots$: an AP of $20$ terms.

$$\alpha=\frac{20}{2}\left(14+19\cdot10\right)=2040$$

Step 2. The root

$\frac{\alpha}{1020}=2$, so $\tan^2\beta$ is a root of $(x-1)(x+2)=0$. It is positive, so $\tan^2\beta=1$ and $\beta=\frac\pi4$.

Step 3. Value

$$\frac12+3\cdot\frac12=2$$

Answer: option (1)

Example 59 (JEE Advanced 2013, Paper 1). Let $S_n=\displaystyle\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2$. Then $S_n$ can take value(s)

▸ SHOW SOLUTION

Step 1. Sign pattern

$\frac{k(k+1)}{2}$ for $k=1,2,3,4$ is $1,3,6,10$, so the signs run $-,-,+,+$ and repeat every four terms.

Step 2. One block of four

For $k=4m+1$ to $4m+4$:

$$(4m+3)^2-(4m+1)^2=2(8m+4)$$$$(4m+4)^2-(4m+2)^2=2(8m+6)$$$$\text{sum of the block}=32m+20$$

Step 3. Add $n$ blocks

$$S_n=\sum_{m=0}^{n-1}(32m+20)=16n^2+4n$$

Step 4. Test the options

$n=8$ gives $1056$ and $n=9$ gives $1332$. $1088$ and $1120$ fall between and are not of this form.

Answer: (A) and (D)

Example 60 (IITianForum). The sum of the first $20$ terms of the series $1^2+2\cdot2^2+3^2+2\cdot4^2+5^2+2\cdot6^2+\cdots$ is

▸ SHOW SOLUTION

Step 1. Split the weights

Every square appears once, and the even squares appear once more:

$$S=\sum_{k=1}^{20}k^2+\sum_{j=1}^{10}(2j)^2$$

Step 2. Evaluate

$$2870+4\cdot385=2870+1540$$

Answer: $4410$, option (B)

Example 61 (JEE Main 2025, 8 April Shift 2). If $\dfrac{1}{1^4}+\dfrac{1}{2^4}+\dfrac{1}{3^4}+\cdots\infty=\dfrac{\pi^4}{90}$, $\dfrac{1}{1^4}+\dfrac{1}{3^4}+\dfrac{1}{5^4}+\cdots\infty=\alpha$, $\dfrac{1}{2^4}+\dfrac{1}{4^4}+\dfrac{1}{6^4}+\cdots\infty=\beta$, then $\dfrac\alpha\beta$ is equal to

▸ SHOW SOLUTION

Step 1. The even part is a multiple of the whole

$$\beta=\sum\frac{1}{(2k)^4}=\frac{1}{16}\cdot\frac{\pi^4}{90}$$

Step 2. The odd part

$$\alpha=\frac{\pi^4}{90}-\beta=\frac{15}{16}\cdot\frac{\pi^4}{90}$$

Answer: $\frac\alpha\beta=15$, option (3)

Method of differences

If the $k$th term can be written as the difference of two consecutive values of one function,

$$T_k=V_k-V_{k-1}$$

then the sum collapses, every middle value cancelling with its neighbour:

$$\sum_{k=1}^{n}T_k=V_n-V_0$$

This is a telescoping sum. The cut of the unit segment below shows it for $\frac{1}{k(k+1)}=\frac1k-\frac1{k+1}$: the pieces fill $[0,1)$ and the first $n$ of them reach $1-\frac{1}{n+1}$.

The unit segment cut into pieces of lengths one over 1 times 2, one over 2 times 3, and so on, with cut points one half, two thirds, three quarters

Standard splittings

The fourth row rests on a factorisation used again and again:

$$k^4+k^2+1=(k^2-k+1)(k^2+k+1)$$

and $(k+1)^2-(k+1)+1=k^2+k+1$, so the second factor at $k$ is the first factor at $k+1$.

Finding $T_n$ from differences of terms

When the terms themselves are not given by a formula, look at the differences of consecutive terms.

A difference table for 1, 3, 11, 25, 45, 71 with first differences 2, 8, 14, 20, 26 and constant second differences 6

Then find the constants from the first few terms. The same idea in one line: if $T_n-T_{n-1}=t_n$ for $n\ge2$, then

$$T_n=T_1+\sum_{k=2}^{n}t_k$$

Examples

Example 62 (JEE Main 2026, 5 April Shift 1). $\displaystyle\sum_{n=1}^{10}\frac{528}{n(n+1)(n+2)}$ is equal to

▸ SHOW SOLUTION

Step 1. Telescope

$$\sum_{n=1}^{10}\frac{1}{n(n+1)(n+2)}$$$$=\frac12\left[\frac{1}{1\cdot2}-\frac{1}{11\cdot12}\right]$$

Step 2. Evaluate

$$\frac12\cdot\frac{65}{132}=\frac{65}{264}$$

Answer: $528\cdot\frac{65}{264}=130$, option (2)

Example 63 (JEE Main 2026, 28 Jan Shift 2). If $\displaystyle\sum_{r=1}^{25}\frac{r}{r^4+r^2+1}=\frac pq$, where $p,q$ are positive integers with $\gcd(p,q)=1$, then $p+q$ is equal to ______.

▸ SHOW SOLUTION

Step 1. Split the term

$$\frac{2r}{r^4+r^2+1}=\frac{1}{r^2-r+1}-\frac{1}{r^2+r+1}$$

Step 2. Telescope

$$\sum_{r=1}^{25}=\frac12\left[1-\frac{1}{651}\right]=\frac{325}{651}$$

$325=5^2\cdot13$ and $651=3\cdot7\cdot31$ share no factor.

Answer: $p+q=976$

Example 64 (JEE Main 2026, 5 April Shift 2). If the sum of the first $10$ terms of the series $\dfrac{1}{1+1^4\cdot4}+\dfrac{2}{1+2^4\cdot4}+\dfrac{3}{1+3^4\cdot4}+\cdots$ is $\dfrac mn$, $\gcd(m,n)=1$, then $m+n$ is equal to

▸ SHOW SOLUTION

Step 1. Factor the denominator

$$1+4r^4=(2r^2+1)^2-(2r)^2$$$$=(2r^2-2r+1)(2r^2+2r+1)$$

Step 2. Split

The two factors differ by $4r$, so

$$\frac{4r}{1+4r^4}=\frac{1}{2r^2-2r+1}-\frac{1}{2r^2+2r+1}$$

and $2(r+1)^2-2(r+1)+1=2r^2+2r+1$, so it telescopes.

Step 3. Sum

$$\frac14\left[1-\frac{1}{221}\right]=\frac{55}{221}$$

Answer: $m+n=276$, option (3)

Example 65 (IITianForum). If $\displaystyle\sum_{n=1}^{20}\frac{2n+1}{n^2(n+1)^2}=\frac pq$ with $\gcd(p,q)=1$, then $p+q$ is equal to ______.

▸ SHOW SOLUTION

Step 1. Split

$(n+1)^2-n^2=2n+1$, so

$$\frac{2n+1}{n^2(n+1)^2}=\frac{1}{n^2}-\frac{1}{(n+1)^2}$$

Step 2. Telescope

$$\sum_{n=1}^{20}=1-\frac{1}{441}=\frac{440}{441}$$

Answer: $p+q=881$

Example 66 (IITianForum). The value of $\displaystyle\sum_{n=3}^{\infty}\ln\left(1-\frac{4}{n^2}\right)$ is

▸ SHOW SOLUTION

Step 1. Turn the sum into a product

$$\sum_{n=3}^{N}\ln\left(1-\frac{4}{n^2}\right)=\ln\prod_{n=3}^{N}\frac{(n-2)(n+2)}{n\cdot n}$$

Step 2. Each half of the product telescopes

$$\prod_{n=3}^{N}\frac{n-2}{n}=\frac{1\cdot2}{(N-1)N}$$$$\prod_{n=3}^{N}\frac{n+2}{n}=\frac{(N+1)(N+2)}{3\cdot4}$$

Step 3. Let $N\to\infty$

The product tends to $\frac{2}{12}=\frac16$, so the sum tends to $\ln\frac16$.

Answer: option (A)

Example 67 (JEE Advanced 2025, Paper 2). Let

$$\alpha=\frac{1}{\sin60^\circ\sin61^\circ}+\frac{1}{\sin62^\circ\sin63^\circ}$$ $$+\cdots+\frac{1}{\sin118^\circ\sin119^\circ}$$

Then the value of $\left(\dfrac{\operatorname{cosec}1^\circ}{\alpha}\right)^2$ is ______.

▸ SHOW SOLUTION

Step 1. Split one term

$\sin1^\circ=\sin\big((x+1^\circ)-x\big)$, so

$$\frac{\sin1^\circ}{\sin x\sin(x+1^\circ)}=\cot x-\cot(x+1^\circ)$$

Step 2. The whole sum

$$\alpha\sin1^\circ=\cot60^\circ-\cot61^\circ+\cot62^\circ$$$$\qquad-\cot63^\circ+\dots-\cot119^\circ$$

Step 3. Cancel through $\cot(180^\circ-x)=-\cot x$

The term at $k^\circ$ carries the sign $(-1)^k$, and $k$ and $180-k$ have the same parity. So the terms at $61^\circ$ and $119^\circ$ cancel, those at $62^\circ$ and $118^\circ$ cancel, and so on up to $89^\circ$ and $91^\circ$. Left are $\cot60^\circ$ and $\cot90^\circ=0$:

$$\alpha\sin1^\circ=\frac{1}{\sqrt3}$$

Answer: $\left(\frac{\operatorname{cosec}1^\circ}{\alpha}\right)^2=3$

Example 68 (JEE Main 2025, 3 April Shift 1). The sum $1+3+11+25+45+71+\cdots$ up to $20$ terms is equal to

▸ SHOW SOLUTION

Step 1. Differences

The first differences $2,8,14,20,26$ form an AP, so $T_n=an^2+bn+c$.

Step 2. Constants

From $T_1=1$, $T_2=3$, $T_3=11$: $a=3$, $b=-7$, $c=5$.

$$T_n=3n^2-7n+5$$

Step 3. Sum

$$3(2870)-7(210)+100=7240$$

Answer: option (1)

Example 69 (IITianForum). A sequence $a_1,a_2,\ldots$ has $a_1=5$ and $a_n-a_{n-1}=n^2$ for every $n\ge2$. If $a_k=389$, then $k$ is equal to

▸ SHOW SOLUTION

Step 1. Add the differences

$$a_k=5+\sum_{n=2}^{k}n^2=4+\frac{k(k+1)(2k+1)}{6}$$

Step 2. Solve

$$\frac{k(k+1)(2k+1)}{6}=385=\frac{10\cdot11\cdot21}{6}$$

Answer: $k=10$, option (C)

Example 70 (JEE Advanced 2022, Paper 1). Let $a_1,a_2,a_3,\ldots$ be an arithmetic progression with $a_1=7$ and common difference $8$. Let $T_1,T_2,T_3,\ldots$ be such that $T_1=3$ and $T_{n+1}-T_n=a_n$ for $n\ge1$. Then, which of the following is/are TRUE?

▸ SHOW SOLUTION

Step 1. Find $T_n$

$a_k=8k-1$, so

$$T_n=3+\sum_{k=1}^{n-1}(8k-1)=4n^2-5n+4$$

Step 2. Terms

$T_{20}=1504$ and $T_{30}=3454$.

Step 3. Sums

$$\sum_{k=1}^{20}T_k=4(2870)-5(210)+80=10510$$$$\sum_{k=1}^{30}T_k=37820-2325+120$$$$=35615$$

Answer: (B) and (C)

Sequences given by a recurrence

A recurrence gives each term from the earlier ones, for example $a_{n+2}=5a_{n+1}-6a_n$. Three patterns cover most JEE questions.

A sum like $\sum a_nx^n$ for a recurrence is found by the same multiply and subtract method used for the AGP: write $S$, $xS$, $x^2S$ and use the recurrence to cancel.

Examples

Example 71 (JEE Main 2025, 28 Jan Shift 1). Let $\langle a_n\rangle$ be a sequence such that $a_0=0$, $a_1=\dfrac12$ and $2a_{n+2}=5a_{n+1}-3a_n$, $n=0,1,2,3,\dots$. Then $\displaystyle\sum_{k=1}^{100}a_k$ is equal to

▸ SHOW SOLUTION

Step 1. Differences form a GP

$$2(a_{n+2}-a_{n+1})=3(a_{n+1}-a_n)$$

So $a_{n+1}-a_n=\frac12\left(\frac32\right)^n$.

Step 2. Find $a_n$

$$a_n=\sum_{k=0}^{n-1}\frac12\left(\frac32\right)^k=\left(\frac32\right)^n-1$$

Step 3. Sum

$$\sum_{k=1}^{100}a_k=3\left[\left(\frac32\right)^{100}-1\right]-100$$$$=3a_{100}-100$$

Answer: option (2)

Example 72 (IIT JEE 2010, Paper 2). Let $a_1,a_2,a_3,\ldots,a_{11}$ be real numbers satisfying $a_1=15$, $27-2a_2>0$ and $a_k=2a_{k-1}-a_{k-2}$ for $k=3,4,\ldots,11$. If $\dfrac{a_1^2+a_2^2+\cdots+a_{11}^2}{11}=90$, then the value of $\dfrac{a_1+a_2+\cdots+a_{11}}{11}$ is equal to ______.

▸ SHOW SOLUTION

Step 1. The recurrence gives an AP

$a_k-a_{k-1}=a_{k-1}-a_{k-2}$, so $a_k=15+(k-1)d$.

Step 2. Sum of squares

$$\sum_{i=0}^{10}(15+id)^2=990$$$$7d^2+30d+27=0$$

So $d=-3$ or $d=-\frac97$.

Step 3. Use the condition

$27-2a_2>0$ means $a_2<13.5$, so $d<-1.5$ and $d=-3$.

Step 4. Mean

The mean of an AP of $11$ terms is its middle term $a_6=15-15$.

Answer: $0$

Example 73 (JEE Advanced 2019, Paper 1). Let $\alpha$ and $\beta$ be the roots of $x^2-x-1=0$, with $\alpha>\beta$. For all positive integers $n$, define $a_n=\dfrac{\alpha^n-\beta^n}{\alpha-\beta}$, $n\ge1$, $b_1=1$ and $b_n=a_{n-1}+a_{n+1}$, $n\ge2$. Then which of the following options is/are correct?

▸ SHOW SOLUTION

Step 1. The recurrence

$\alpha^2=\alpha+1$ and $\beta^2=\beta+1$ give $a_{n+2}=a_{n+1}+a_n$, with $a_1=a_2=1$.

Step 2. Option (A)

$a_k=a_{k+2}-a_{k+1}$, so the sum telescopes to $a_{n+2}-a_2=a_{n+2}-1$. True.

Step 3. Option (B)

Let $S=\sum a_nx^n$ with $x=\frac1{10}$. Using the recurrence,

$$S-xS-x^2S=a_1x=x$$$$S=\frac{0.1}{0.89}=\frac{10}{89}$$

True.

Step 4. Option (C)

With $\alpha\beta=-1$, $\alpha+\frac1\alpha=\alpha-\beta$ and $\beta+\frac1\beta=\beta-\alpha$. Then

$$a_{n-1}+a_{n+1}=\frac{\alpha^n(\alpha-\beta)+\beta^n(\alpha-\beta)}{\alpha-\beta}$$$$=\alpha^n+\beta^n$$

and $b_1=1=\alpha+\beta$. True.

Step 5. Option (D)

$\sum\frac{\alpha^n}{10^n}+\sum\frac{\beta^n}{10^n}=\frac{\alpha}{10-\alpha}+\frac{\beta}{10-\beta}$, which works out to $\frac{12}{89}$. False.

Answer: (A), (B) and (C)

Example 74 (JEE Main 2026, 21 Jan Shift 1). Let $a_1=1$ and for $n\ge1$, $a_{n+1}=\dfrac12a_n+\dfrac{n^2-2n-1}{n^2(n+1)^2}$. Then $\left\lvert\displaystyle\sum_{n=1}^{\infty}\left(a_n-\frac{2}{n^2}\right)\right\rvert$ is equal to ______.

▸ SHOW SOLUTION

Step 1. Split the extra term

$$\frac{2}{(n+1)^2}-\frac{1}{n^2}=\frac{n^2-2n-1}{n^2(n+1)^2}$$

Step 2. Rewrite the recurrence

With $b_n=a_n-\frac{2}{n^2}$:

$$b_{n+1}+\frac{2}{(n+1)^2}=\frac12b_n+\frac{1}{n^2}$$$$\qquad+\frac{2}{(n+1)^2}-\frac{1}{n^2}$$

So $b_{n+1}=\frac12b_n$, a GP.

Step 3. Sum the GP

$b_1=1-2=-1$, so $b_n=-\left(\frac12\right)^{n-1}$ and $\sum b_n=-2$.

Answer: $2$

Bounding a sum

When a sum has no closed form, it can still be trapped between two sums that do. If every term lies between $m$ and $M$, the sum of $n$ terms lies between $nm$ and $nM$. When both bounds tend to the same limit, so does the sum (the sandwich theorem).

Example 75. The value of $\displaystyle\lim_{n\to\infty}\sum_{k=1}^{n}\frac{n}{n^2+k}$ is

▸ SHOW SOLUTION

Step 1. Bound each term

For $1\le k\le n$:

$$\frac{n}{n^2+n}\le\frac{n}{n^2+k}\le\frac{n}{n^2+1}$$

Step 2. Bound the sum

$$\frac{n^2}{n^2+n}\le\sum_{k=1}^{n}\frac{n}{n^2+k}\le\frac{n^2}{n^2+1}$$

Step 3. Limit

Both bounds tend to $1$.

Answer: option (C)

Which method to use

What JEE asks

JEE Main asks one or two questions from this chapter in nearly every shift. In 2026 the most frequent types were: conditions on an AP solved with $a$ and $d$ (Examples 4 to 8 and 13), products and sums of GP terms (Examples 23 to 31), AP and GP linked in one question (Examples 39 to 41), sums whose general term is a polynomial (Examples 52 to 54 and 56 to 58), and telescoping fractions with $k^4+k^2+1$ or $1+4k^4$ in the denominator (Examples 62 to 64). Two 2026 shifts asked for a quadratic built from a sum of products whose roots differ by $2$ (Examples 56 and 57).

JEE Advanced asks fewer but deeper questions: an AP carried into a GP by $f(x+y)=f(x)f(y)$ (Example 34) and a trigonometric telescoping sum (Example 67) in 2025, coefficients in AP with integer roots (Example 22) in 2026, and in earlier years common terms of two APs, the sign pattern $(-1)^{k(k+1)/2}$, recurrences of Fibonacci type and sums defined by differences (Examples 15, 59, 70 and 73).

Quick recall

Practise these methods on real exam questions in the doMath Sequence and Series chapter. Maxima and minima by $A\ge G\ge H$ are developed in the Relation Between AM, GM and HM note, and quadratic equations, which appear in several examples here, in the Quadratic Equations note.