Q34 / 75 · JEE Main 2026 (28 January Shift 1) · Mathematics

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

  • A
    16
  • B
    21
  • Correct answer
    C
    19
  • D
    24

Solution

Let $d$ and $d'$ be the common differences of the two APs respectively.
Given: $d = d' + 13$.
From the second AP: $b_{43} - b_{31} = 12d'$, so $-385 - (-277) = 12d'$, giving $-108 = 12d'$ and $d' = -9$.
Therefore $d = -9 + 13 = 4$.
For the first AP: $a_{78} = a_1 + 77d$, so $327 = a_1 + 77(4) = a_1 + 308$.
Thus $a_1 = 19$

Hence, the correct answer is option: 'C'.

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