Question

The sum of four consecutive terms of an A.P is 32 & the ration of the product of first and last terms to the product of two middle terms is 7:15. Find the number. ​

Answer 1

Suppose the four consecutive numbers are (a - 3d), (a - d), (a + d) and (a + 3d)

Thus, a-3d + a - d + a + d + a + 3d = 32

4a = 32

a = 8 ......(1st eq)

Now, (a - 3d)(a + 3d)/(a - d)(a + d) = 7/15

15(a² - 9d²) = 7(a² - d²)

15a² - 135d² = 7a² - 7d²

15a² - 7a² = 135d² - 7d² 

8a² = 128d²..... (2nd eq)

Put the value of a = 8 in second equation we get -

8(8)² = 128d²

128d² = 512

d² = 512/128

d² = 4 = 2

So, the four consecutive numbers are

8 - (3*2) = 8 - 6 = 2

8 - 2 = 6

8 + 2 = 10

8 + (3*2)

8 + 6 = 14

Four consecutive numbers are 2, 6, 10 and 14