Question:-

An AP consists of 37 TERMS . The SUM of first three terms is 12 and sum of last three terms is 318. Find first and last terms.

Answer:-

Given:-

Number of terms in an AP (n) = 37

Sum of first three terms = 12.

First three terms of an AP are a , a + d , a + 2d where a is the first term and d is common difference.

So,

⟹ a + a + d + a + 2d = 12

⟹ 3a + 3d = 12

⟹ 3(a + d) = 12

⟹ a + d = 12/3

⟹ a + d = 4 -- equation (1).

Also given that,

Sum of last three terms = 318.

That is;

⟹ 35th term + 36th term + 37TH term = 318.

We know that,

nth term of an AP (aₙ) = a + (n - 1)d

So,

⟹ a + (35 - 1)d + a + (36 - 1)d + a + (37 - 1)d = 318

⟹ 3a + 34d + 35d + 36d = 318

⟹ 3a + 105d = 318

⟹ 3(a + 35d) = 3 × 106

⟹ a + 35d = 106 -- equation (2)

Now,

Subtract equation (1) from (2).

⟹ a + 35d - (a + d) = 106 - 4

⟹ a + 35d - a - d = 102

⟹ 34d = 102

⟹ d = 102/34

⟹ d = 3

Substitute d = 3 in equation (1).

⟹ a + 3 = 4

⟹ a = 4 - 3

⟹ a = 1

Hence,

Last term = 37th term = a + 36d

⟹ a₃₇ = 1 + 36(3)

⟹ a₃₇ = 1 + 108

⟹ a₃₇ = 109

∴

• First term of the AP = a = 1

• Last term of the AP = a₃₇ = 109.