Question

The sum of 4 consecutive number in an AP is 32 and the ratio of the product of the first and the last term to the product of two middle term is 7:15 arrange the numbers

Answer 1

Given :-

• The sum of 4 consecutive number in an AP is 32 and the ratio of the product of the first and the last term to the product of two middle term is 7:15.

Solution:-

Let the four consecutive numbers in AP be (a - 3d), (a - d), (a + d) and (a + 3d)

So,

According to the question.

➱  a-3d + a - d + a + d + a + 3d = 32

➱  4a = 32

➱  a = 32/4

➱  a = 8 ....(1)

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

Putting the value of a = 8 in above we get.

➪  8(8)² = 128d²

➪  128d² = 512

➪  d² = 512/128

➪  d² = 4

➪  d = 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