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

The sum of four consecutive numbers in an AP is 32 and the ratio of th

1.	The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last term to the product of the middle terms is 7:15. Find the numbers.
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 = 32/4 a = 8 ......(1) Now, (a - 3d)(a + 3d)/(a - d)(a + d) = 7/15 15(a² - 9d²) = 7(a² - d²) 15a² - 135d² = 7a² - 7d² 15a² - 7a² = 135d² - 7d² 8a² = 128d² Putting the value of a = 8 in above we get. 8(8)² = 128d² 128d² = 512 d² = 512/128 d² = 4 d = 2 So, the four consecutive numbers are 8 - (32) 8 - 6 = 2 8 - 2 = 6 8 + 2 = 10 8 + (32) 8 + 6 = 14 Four consecutive numbers are 2, 6, 10 and 14

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

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