Question

35. An arithmetic progression consists of 37 terms. The sum of the first 3 terms of
it is 12 and the sum of its last 3 terms is 318, then find the first and last terms
of the progression​

Answer 1

Step-by-step explanation:

Let the A.P. be:

a, a+d, a+2d, a+3d ... a+(36-1)d, a + (37-1)d

According to question,

a + a + d + a + 2d = 12

=> 3a + 3d = 12

=> a + d = 4

=> d = 4 - a --- 1

Also,

a + 34d + a + 35d + a + 36d = 318

=> 3a + 105d = 318

=> a + 35d = 106

From equation 1

=> a + 35(4-a) = 106

=> 35×4 - 35a + a = 106

=> 140 - 34a = 106

=> 34a = 34

=> a = 1

d = 4 - a = 3

Therefore,

First term is a = 1

And last term is a + 36d = 1 + 36×3 = 109