Question

the sum of 4 consecutive terms which are in an AP is 32 and the ratio of product of first & last term of the product of two middle terms is 7:15. find the number​

Answer 1

Answer

Let the four consecutive numbers in AP be;

(a - 3d), (a - d), (a + d) and (a + 3d)

So, according to the question

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

4a = 32

a = 8

Now,

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

15(a² - 9d²) = 7(a² - d²)

15a² - 135d² = 7a² - 7d²

8a² = 128d²

Putting the value of a = 8, we get

d² = 4

d = ±2

So, the four consecutive numbers are 2, 6, 10, 14 or 14, 10, 6, 2

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Answer 2

Step-by-step explanation:

Let 4 consecutive terms of the A.P. be (a - 3d), (a - d), (a + d), (a + 3d)

Therefore,

According to the problem,

Sum of four consecutive terms which are in A.P. is 32

So,

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

=> 4a = 32

=> a = 8

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Again,

ratio of product of first & last term of the product of two middle terms is 7:15

So,

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

So, the terms are

• (a - 3d) = (8 - 6) = 2
• (a - d) = (8 - 2) = 6
• (a + d) = (8 + 2) = 10
• (a + 3d) = (8 + 6) = 14

_____________________________

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