Question

the sum of four consecutive nubers in an ap is 32 and the ratio of the product of the first and the last term to the product of two muddle most terms is 7:15. find the numbers​

Answer 1

Answer:

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Answer 2

Answer:

The Require No of AP are 5 , 7 , 9 , 11.

Step-by-step explanation:

Given: Sum of four consecutive term of an AP = 32

To find: 4 consecutive numbers

let 4 terms of AP are a - 3d , a - d , a + d & a + 3d

According to question,

( a - 3d ) + ( a - d ) + ( a + d ) + (a + 3d ) = 32

4a = 32

a = 8

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

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

$\frac{8^2-9d^2}{8^2-d^2}=\frac{7}{15}$

$15\times(64-9d^2)=7\times(64-d^2)$

$512-72d^2=448-8d^2$

$72d^2-8d^2=512-448$

$64d^2=64$

$d=\pm1$

when d = 1

1st term = a - 3d = 8 - 3 = 5

2nd term = a - d = 8 - 1 = 7

3rd term = a + d = 8 + 1 = 9

4th term = a + 3d = 8 + 3 = 11

when d = -1

1st term = a - 3d = 8 + 3 = 11

2nd term = a - d = 8 + 1 = 9

3rd term = a + d = 8 - 1 = 7

4th term = a + 3d = 8 - 3 = 5

Therefore, The Require No of AP are 5 , 7 , 9 , 11.