Sequences & SeriesmediumPYQ · JEE Main · 6 Apr 2026 · Shift 1 (Morning)Free

AP and GP Linked Conditions: Sum = 34/9 | JEE Main 2026

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

Question
The sum of the first ten terms of an A.P. is 160160 and the sum of the first two terms of a G.P. is 88. 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
A349\dfrac{34}{9}correct
B3413\dfrac{34}{13}
C329\dfrac{32}{9}
D3213\dfrac{32}{13}
Solution
Step 1: Let the A.P. have first term aa, common difference dd. Sn=n2(2a+(n−1)d)S_n=\dfrac{n}{2}\big(2a+(n-1)d\big):
S10=5(2a+9d)=160⇒2a+9d=32.(∗)S_{10}=5(2a+9d)=160\Rightarrow2a+9d=32.\qquad(\ast)
 Step 2: G.P. is d, da, da2,…d,\ da,\ da^2,\ldots
d+da=8⇒d(1+a)=8.(∗∗)d+da=8\Rightarrow d(1+a)=8.\qquad(\ast\ast)
 Step 3: From (∗)(\ast), a=32−9d2a=\dfrac{32-9d}{2}. Into (∗∗)(\ast\ast):
d(1+32−9d2)=8⇒d⋅34−9d2=8⇒d(34−9d)=16.d\left(1+\dfrac{32-9d}{2}\right)=8\Rightarrow d\cdot\dfrac{34-9d}{2}=8\Rightarrow d(34-9d)=16.
⇒34d−9d2=16⇒9d2−34d+16=0.\Rightarrow34d-9d^2=16\Rightarrow9d^2-34d+16=0.
 Step 4: ∴d1+d2=349\therefore d_1+d_2=\dfrac{34}{9}. Correct answer: (1)
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