JEE Main 2026 · 6 Apr · Shift 1 · Mathematics Q6

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MathematicsSingle correct+4 / -1Question 6 of 75

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

\dfrac{34}9

\dfrac{34}{13}

\dfrac{32}9

\dfrac{32}{13}

Answer and worked solution

Official answer: A

A.P. condition. First term a, difference d: $5(2a+9d)=160$ , so $2a+9d=32$ .

G.P. condition. First term d, ratio a: $d(1+a)=8$ .

Eliminate a.
$d\left(1+\frac{32-9d}2\right)=8\ \Rightarrow\ 9d^2-34d+16=0.$

Sum of the values of d: $\dfrac{34}9$ .

(A)

Topics:Arithmetic progression Geometric progression Roots and coefficients

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JEE Main 2026 · 6 Apr · Shift 1, Mathematics question 6