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Sequences & Series: Match Column Column

JEE Maths question with a full step-by-step solution.

Question

Match Column-I with Column-II.

List-I

AIf

p^{\text{th}}

q^{\text{th}}

and

r^{\text{th}}

terms of a G.P. (common ratio

\ne 1

) are in G.P., then

p, q, r

will be in

B If

a, b, c, d

are non-zero real numbers such that

\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right) \le (ab+bc+cd)^{2}

and

c(a-b) = a(b-c)

, then

a, b, c, d

will be in

CIf

a, b, c, d

are four different positive numbers in G.P., then

\dfrac{a\left(2\log a + \log b\right)}{b},\ \dfrac{b\left(2\log b + \log c\right)}{c},\ \dfrac{c\left(2\log c + \log d\right)}{d}

will be in

DIf

a_1, a_2, a_3

are in A.P.,

a_2, a_3, a_4

are in G.P. and

a_3, a_4, a_5

are in H.P., then

a_1, a_3, a_5

are always in

List-II

PArithmetic progression

Q Geometric progression

R Harmonic progression

S Not a harmonic progression

T A progression whose all the terms are identical

A(A)

\to

(R,S)

B(B)

\to

(P,Q,,T)

C(C)

\to

(P,Q,S)

D(D)

\to

(Q,S)correct

Solution

PART (A) Step 1: Write the three terms of the G.P. With first term

A

and common ratio

R

T_p = AR^{\,p-1}, \qquad T_q = AR^{\,q-1}, \qquad T_r = AR^{\,r-1} .

Step 2: Impose that these are themselves in G.P., i.e.

T_q^{2} = T_p T_r

A^{2}R^{\,2q-2} = A^{2}R^{\,p+r-2} .

Step 3: Cancel

A^{2}

and compare exponents. Since

R \ne 1

, equal powers force equal exponents:

2q - 2 = p + r - 2 \implies 2q = p + r ,

which is exactly the condition for

p, q, r

to be in A.P. Step 4: A non-constant A.P. is never an H.P., so (A) matches (P) and (S). PART (B) Step 5: Handle the inequality with an algebraic identity (this is Lagrange's identity, and can be checked by simply expanding both sides):

\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right) - (ab+bc+cd)^{2} = \left(ac-b^{2}\right)^{2} + (ad-bc)^{2} + \left(bd-c^{2}\right)^{2} .

Step 6: The right side is a sum of squares, so it is always

\ge 0

. The question states the reverse inequality, so the difference must be exactly

0

, forcing each square to vanish:

b^{2} = ac, \qquad bc = ad, \qquad c^{2} = bd .

These say precisely that

a, b, c, d

are in G.P. Step 7: Now use the second condition.

c(a-b) = a(b-c) \implies ca - cb = ab - ac \implies 2ac = ab + bc \implies b = \frac{2ac}{a+c} ,

which says

a, b, c

are in H.P. Step 8: Combine. From Step 6,

b^{2} = ac

; substituting into Step 7,

b = \frac{2b^{2}}{a+c} \implies a + c = 2b ,

a, b, c

are also in A.P. A set of numbers in both A.P. and G.P. must be equal, so

a = b = c

, and then

c^{2} = bd

gives

d = c

as well. Step 9: All four numbers are identical, so the list is simultaneously an A.P., a G.P. and an H.P. Hence (B) matches (P), (Q), (R) and (T). PART (C) Step 10: Write the G.P. as

a,\ b = ar,\ c = ar^{2},\ d = ar^{3}

with

r > 0

r \ne 1

. Step 11: Simplify the first entry. Using

2\log a + \log b = \log\left(a^{2}b\right)

\frac{a}{b}\log\left(a^{2}b\right) = \frac{a}{ar}\log\left(a^{2}\cdot ar\right) = \frac1r \log\left(a^{3}r\right).

Step 12: Do the same for the second and third entries.

\frac{b}{c}\log\left(b^{2}c\right) = \frac1r \log\left(a^{2}r^{2}\cdot ar^{2}\right) = \frac1r\log\left(a^{3}r^{4}\right),

\frac{c}{d}\log\left(c^{2}d\right) = \frac1r \log\left(a^{2}r^{4}\cdot ar^{3}\right) = \frac1r\log\left(a^{3}r^{7}\right).

Step 13: Expand each logarithm as

\log a^{3} + (\text{power})\log r

\frac1r\left[\log a^{3} + \log r\right], \quad \frac1r\left[\log a^{3} + 4\log r\right], \quad \frac1r\left[\log a^{3} + 7\log r\right].

Step 14: Consecutive entries differ by the constant

\dfrac{3\log r}{r}

, so they form an A.P., and (being non-constant, as

r \ne 1

) not an H.P. So (C) matches (P) and (S). PART (D) Step 15: Write down the three conditions.

2a_2 = a_1 + a_3 \ \text{(A.P.)}, \qquad a_3^{2} = a_2a_4 \ \text{(G.P.)}, \qquad a_4 = \frac{2a_3a_5}{a_3+a_5} \ \text{(H.P.)} .

Step 16: Put the H.P. expression for

a_4

into the G.P. relation

a_3^{2} = a_2 \cdot \frac{2a_3a_5}{a_3+a_5} \implies a_3\left(a_3+a_5\right) = 2a_2a_5 .

Step 17: Replace

2a_2

a_1 + a_3

from the A.P. relation.

a_3^{2} + a_3a_5 = \left(a_1+a_3\right)a_5 = a_1a_5 + a_3a_5 .

Step 18: Cancel

a_3a_5

from both sides.

a_3^{2} = a_1a_5 ,

which says

a_1, a_3, a_5

are in G.P. (and not in H.P. in general). So (D) matches (Q) and (S). Answer: (A)

\to

(P), (S); (B)

\to

(P), (Q), (R), (T); (C)

\to

(P), (S); (D)

\to

(Q), (S).

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