Question

the sum of four numbers in an A.P. is 32 and the ratio of the product of the extremes to the product of meams is 7:15. find the number

Answer 1

Hi,

Let a -3d , a - d , a + d , a + 3d are four terms

in an A.P

according to the problem given,

sum of four numbers = 32

a-3d + a - d + a + d + a + 3d = 32

4a = 32

a = 32/4

a = 8

product of the extremes : product of means

= 7:15

( a - 3d )( a + 3d ) / ( a - d )( a + d ) = 7 : 15

( a² - 9d² ) / ( a² - d² ) = 7 / 15

15( a² - 9d² ) = 7( a² - d² )

15a² - 135d² = 7a² - 7d²

- 135d² + 7d² = 7a² - 15a²

- 128d² = - 8a²

d² = 8a²/128

d² = a² / 16

d² = 8² /16

d² = ( 64 /16 )

d² = 4

d = ± √4

d = ±2

Now ,

i ) a = 8 , d = 2

required numbers are

a - 3d = 8 - 3 × 2 = 2

a - d = 8 - 2 = 6

a + d = 8 + 2 = 10

a + 3d = 8 + 3 × 2 = 14

ii ) if a = 8 and d = -2

then 4 terms are

14 , 10 , 6 , 2

I hope this helps you.

:)