Question

the sum of 4 consecutive terms of an Ap is 32 and the ration of the product of first and last term to the product of two middle term is 7:15. find the number​

Answer 1

HELLO DEAR FRIEND

LET THE AP BE ,

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

atq,

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

=> 4d = 32

$\large{d = \frac{ \large{32}}{ \large{4}} }$

so,

$\large \boxed{d = 8}$

given that,

the ration of the product of first and last term to the product of two middle term is 7:15.

then,

$\large{ \frac{ \large{(a - 3d)(a + 3d)}}{ \large{(a - d)(a + d)}} = \frac{ \large{7}}{ \large{15}} }$

$\huge{ \huge{ \frac{ {a}^{2} - {3d}^{2} }{ {a}^{2} - {d}^{2} } = \frac{7}{15} } }$

$\large{ {7a}^{2} - {7d}^{2} = {15a}^{2} - {15d}^{2} }$

$\large{ {7a}^{2} - {15a}^{2} = {7d}^{2} - {15d}^{2} }$

-8a² = - 8d²

a² = d²

putting the value of d

a² = 8²

a = 8

$\huge \boxed{a = 8}$

$\huge \boxed{8}$

SO,

AP = a - 3d, a - d, a + d, a + 3d

= 8 - 3(8) , 8 - 8 , 8 + 8 , 8 + 3(8)

$\huge \underline{AP } =$ $\huge\boxed{16,0,16,32}$

HOPE IT HELPS YOU DEAR FRIEND THANKS