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Mathematics · Sequence and Series

JEE Main 2026 — 6 April, Morning Shift — Question 26

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 8.8. 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 common difference of the A.P., then the sum of all possible values of the first term of the G.P. is:

  1. Option A:

    349\frac{34}{9}

    Correct
  2. Option B:

    3413\frac{34}{13}

  3. Option C:

    329\frac{32}{9}

  4. Option D:

    3213\frac{32}{13}

Answer: A

Step-by-step solution

Sum of 1010 terms of A.P, S10=160\mathrm{S}_{10}=160

5[2a+9d]=1602a+9d=32 G.P. ⇒d,da,da2…⇒d+da=8⇒d+d(32−9d)2=8⇒2d+32d−9d2=8⇒9d2−34d+8=0d1+d2=349\begin{aligned} & 5[2 a+9 d]=160 \\& 2 a+9 d=32 \\& \text { G.P. } \Rightarrow d, d a, d a^{2} \ldots \\& \Rightarrow d+d a=8 \\& \Rightarrow \frac{d+d(32-9 d)}{2}=8 \\& \Rightarrow 2 d+32 d-9 d^{2}=8 \\& \Rightarrow 9 d^{2}-34 d+8=0 \\& d_{1}+d_{2}=\frac{34}{9} \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
6 April, Morning Shift
Subject
Mathematics
Chapter
Sequence and Series
Topic
Arithmetic Progression

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