Question

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

Answer 1

HELLO DEAR,

Let the four consecutive numbers in AP be a-3d,a-d,a+d,a+3d.

THEN,

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

 
=> 4a=32


=>   a=8 1

now,

$\frac{(a - 3d)(a + 3d)}{(a - d)(a + d)} = \frac{7}{15} \\ \\ = > \frac{ {a}^{2} - 9 {d}^{2} }{ {a}^{2} - {d}^{2} } = \frac{7}{15} \\ \\ = > 15( {a}^{2} - 9 {d}^{2} ) = 7( {a}^{2} - {d}^{2} ) \\ \\ = > 15 {a}^{2} - 135 {d}^{2} = 7 {a}^{2} - 7 {d}^{2} \\ \\ = > 15 {a}^{2} - 7 { a }^{2} = - 7 {d}^{2} + 135 {d}^{2} \\ \\ = > 8 {a}^{2} = 128 {d}^{2} \\ \\ = > {a}^{2} = 16 {d}^{2} \\ \\ = > {8}^{2} = 16{d}^{2} \\ \\ = > {d}^{2} = 4 \\ \\ = > d = + - 2$


So,

when a=8 and d=2,

the numbers are 2,6,10,14.

AND,


When a=8, d= -2,

the numbers are 14,10,6,2

Answer 2

The 4 numbers are 14 , 10 , 6 , 2 .

Let the 4 terms be a - 3 d , a - d , a + d , a + 3 d

Given :

The sum of the terms = 32 .

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

= 4 a = 32

= a = 32/4

= a = 8

So a = 8

Given :

ratio of product of first and last term to middle terms = 7 : 15

So :

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

= >  ( a² - 9 d² ) : ( a² - d²  ) = 7 : 15

put a = 8 :

= > ( 64 - 9 d² )/( 64 - d² ) = 7/15

= > 15 ( 64 - 9 d² ) = 7 ( 64 - d² )

= > 960 - 135 d² = 448 - 7 d²

= > 128 d² = 960 - 448

= > 128 d² = 512

= > d² = 512/128

= > d² = 4

= > d = ± 2

The numbers will be :

8 + 6 , 8 + 2 , 8 - 2 , 8 - 6

= 14 , 10 , 6 , 2

The 4 numbers are 14 , 10 , 6 , 2 .