JEE Main 2026 — Jan 28, Shift 1 — Mathematics solution

JEE Main2026Jan 28, Shift 1Mathematics

The common difference of the A.P.: $a_{1}, a_{2}, \dots, a_{m}$ is 13 more than the common difference of the A.P.: $b_{1}, b_{2}, \dots, b_{n}$ . If $b_{31} = - 277$ , $b_{43} = - 385$ and $a_{78} = 327$ , then $a_{1}$ is equal to

A
21
B
24
C
19
D
16

Solution

Let the common differences be $d_{1}$ and $d_{2}$ , with $d_{1} = d_{2} + 13$ .

From the ${b_{n}}$ : $b_{43} - b_{31} = 12 d_{2} = - 385 - (- 277) = - 108$ , so $d_{2} = - 9$ . Hence $d_{1} = 4$ .

For ${a_{n}}$ : $a_{78} = a_{1} + 77 d_{1} = a_{1} + 308 = 327$ , so $a_{1} = 19$ .

Option (3).

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