Question

The sum of four consecutive terms of an AP is 32 and the ratio of the product of the first and the last term to the product of two middle terms is 7 : 15 . Find the numbers

Answer 1

HEYA !

The answer is in the attachment. Please refer to it once!

Hope it helps!

Answer 2

Answer:

 2 , 6, 10 , 14 .

Step-by-step explanation:

Let four consecutive number as

a - 3 d , a - d , a + d , a + 3 d .

Given their sum is 32 .

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

4 a = 32

a = 8 .

Now :

Also given the ratio of the product of the first and the last term to product of two middle term to the product of two middle terms is 7:15 .

$\displaystyle{\dfrac{(a-3d)(a+3d)}{(a-d)(a+d)} = \dfrac{7}{15} }\\\\\\\displaystyle{\dfrac{a^2-(3d)^2}{a^2-d^2} = \dfrac{7}{15} }$

We have a = 8

$\displaystyle{\dfrac{64-(3d)^2}{64-d^2} = \dfrac{7}{15} }\\\\\\\displaystyle{\dfrac{64-9d^2}{64-d^2} = \dfrac{7}{15} }$

960 - 135 d² = 448 - 7 d²

135 d² - 7 d²  = 960 - 448

128 d²  = 512

d²  = 4

d = ± 2

When d = 2

Numbers are ,

a - 3 d , a - d ,  a + d , a + 2 d

= > 8 - 6 = 2

= > 8 - 2 = 6

= > 8 + 2 = 10

= > 8 + 6 = 14

When d = - 2

= > 8 + 6 = 14

= > 8 + 2 = 10

= > 8 - 2 = 6

= > 8 - 6 = 2 .

But both ways we get same numbers.

i.e.  2 , 6, 10 , 14 .