Question

The number of terms of AP:3,7,11,15,.... to taken so that sum is 136

Answer 1

$a = 3 \\ d = 4 \\ s = 136 \\let \: n = x \\ s = \frac{n}{2} (2a + (n - 1)d) \\ 136 = \frac{x}{2} (6 + (x - 1)4) \\ \frac{272}{x} = 6 + 4x - 4 \\ 272 = 2x + 4 {x}^{2} \\ 2 {x}^{2} + x - 136 \\ 2 {x}^{2} - 16x + 17x - 136 \\ 2x(x - 8) + 17(x - 8) \\ (2x+ 17)(x - 8) \\ x = \frac{ - 17}{2} \: \: and \: \: 8$
since n cannot be negative
hence 8th term is the required answer