Question

find four consecutive terms in ap whose sum is 88 and the sum of the first and the third terms is 40

Answer 1

Answer:

16, 20, 24, 28

Step-by-step explanation:

Let four consecutive terms are a - 3d , a - d , a + d , a + 3d , where a and d is real numbers.

A/C to question,

sum of all four terms = 88

(a - 3d) + (a - d) + (a + d) + (a + 3d) = 88

4a = 88

a = 22 ......(1)

again, sum of 1st and 3rd term is 40

e.g., (a - 3d) + (a + d) = 40

2a - 2d = 40

a - d = 20

from equation (1),

22 - d = 20

d = 2

hence, a = 22 and d = 2

then, a - 3d = 22 - 3 × 2 = 22 - 6 = 16

a - d = 22 - 2 = 20

a + d = 22 + 2 = 24

a + 3d = 22 + 3× 2 = 22 + 6 = 28

therefore, four consecutive term in an A.P are 16, 20, 24, 28