The fourth term of an A.P. is 11 and the eight term exceeds twice the fourth term by 5. Find the A.P. and the sum of first 50 terms.

Let the first term of an A.P. be a and common difference be d.

Given,

⇒ a₄ = 11

⇒ a + (4 - 1)d = 11

⇒ a + 3d = 11 …….(i)

Also,

⇒ a₈ = 2a₄ + 5

⇒ a + (8 - 1)d = 2[a + (4 - 1)d] + 5

⇒ a + 7d = 2a + 6d + 5

⇒ a - 2a + 7d - 6d = 5

⇒ -a + d = 5 ……..(ii)

Adding (i) and (ii) we get,

⇒ a + 3d + (-a + d) = 11 + 5

⇒ 4d = 16

⇒ d = 4.

Substituting value of d in (i) we get,

⇒ a + 3(4) = 11

⇒ a + 12 = 11

⇒ a = -1.

A.P. = a, (a + d), (a + 2d), ……….

= -1, 3, 7, ………..

$S = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] = \dfrac{50}{2}[2 \times (-1) + (50 - 1) \times 4] \\[1em] = 25[-2 + 196] \\[1em] = 25 \times 194 \\[1em] = 4850.$

Hence, A.P. = -1, 3, 7, ……….. and sum of first 50 terms = 4850.