JEE Main 2026 — Apr 2, Shift 2 — Mathematics solution

JEE Main2026Apr 2, Shift 2Mathematics

Let $a_{1}, a_{2}, a_{3}, \dots$ be an A.P. and $g_{1} = a_{1}, g_{2}, g_{3}, \dots$ be an increasing G.P. If $a_{1} = a_{2} + g_{2} = 1$ and $a_{3} + g_{3} = 4$ , then $a_{10} + g_{5}$ is equal to :

A
$81$
B
$76$
C
$62$
D
$55$

Solution

Let the A.P. have common difference $d$ and the G.P. have common ratio $r$ , with both sequences starting at $1$ .

The condition $a_{2} + g_{2} = 1$ gives $(1 + d) + r = 1$ , so $d + r = 0$ .

The condition $a_{3} + g_{3} = 4$ gives $(1 + 2 d) + r^{2} = 4$ . Substituting $d = - r$ leads to $r^{2} - 2 r - 3 = 0$ , whose roots are $r = 3$ or $r = - 1$ .

Since the G.P. is increasing, $r = 3$ and consequently $d = - 3$ .

Therefore $a_{10} = 1 + 9 d = - 26$ and $g_{5} = r^{4} = 81$ , giving $a_{10} + g_{5} = 55$ .

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