Find the sum of first 51 terms of an A.P. whose second and third terms are 14 and 18, respectively.

Question :Find the sum of first 51 terms of an A.P. whose second and third terms are 14 and 18, respectively.

The correct answer :Let the first term of the AP be a and the common difference be d. Then, the second term is a + d, and the third term is a + 2d.

According to the problem, we have:

a + d = 14 (1) a + 2d = 18 (2)

Subtracting equation (1) from equation (2), we get:

d = 4

Substituting this value of d into equation (1), we get:

a + 4 = 14

a = 10

Therefore, the AP is: 10, 14, 18, 22, …

The sum of the first n terms of an AP is given by the formula:

S_n = n/2 [2a + (n-1)d]

Substituting a = 10, d = 4, and n = 51, we get:

S_51 = 51/2 [2(10) + (51-1)4] = 51/2 [20 + 200] = 51/2 x 220 = 5625

Therefore, the sum of the first 51 terms of the AP is 5625.