JEE Main 2026 Apr 6 Shift 1, Mathematics Q6: Arithmetic, Geometric And Harmonic Progressions

JEE Main2026Apr 6, Shift 1Mathematics

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:

A
$\frac{34}{9}$
B
$\frac{34}{13}$
C
$\frac{32}{9}$
D
$\frac{32}{13}$

Solution

Let the A.P. have first term $a$ and common difference $d$ . Then the G.P. has first term $d$ and common ratio $a$ .

Sum of $10$ A.P. terms: $\frac{10}{2} (2 a + 9 d) = 160 \Rightarrow 2 a + 9 d = 32$ , so $a = \frac{32 - 9 d}{2}$ .

Sum of first two G.P. terms: $d + d a = 8$ , i.e. $d (1 + a) = 8$ .

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

By Vieta's, the sum of the two roots is $d_{1} + d_{2} = \frac{34}{9}$ .

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JEE Main 2026 Apr 6 Shift 1, Mathematics Q6: Arithmetic, Geometric And Harmonic Progressions

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