Question

Which term of the A.P. 3, 14, 25, 36.... is 121 more than its 82th term ?

Answer 1

$an = a82 + 121 \\ a + (n - 1)d = a + 81d + 121 \\ (n - 1) d= 81d + 121 \\ (n - 1) 11 = 81 \times 11 + 121 \\ n - 1 = \frac{11(81 + 11)}{11 } \\ n - 1 = 81 + 11 \\ n = 93$