Question

for four consecutive terms in an AP whose sum is 88 and the sum of 1 and 3 term is 40

Answer 1

Answer :

The terms of the A. P are 16, 20, 24, 28.

Step by step explanation :

Let the four consecutive terms of the A. P be a - 3d, a - d, a + d, a + 3d.

Given,

Sum of first four consecutive terms = 88

a - 3d + a - d + a + d + a + 3d = 88

4a = 88

a = 22

Sum of 1st & 3rd term is 40 .

a - 3d + a + d = 40.

2a - 2d = 40

2(22) - 2d = 40.

44 - 2d = 40.

44 - 40 = 2d

4 = 2d.

d = 2.

Therefore, The terms of the A. P are 16, 20, 24, 28.

Answer 2

Answer:

Step-by-step explanation: