4. The sum of four consecutive numbers in ar AP is 32 and the ratio of the product of the first and the last term tothe product of two middle terms is 7: 15 find the numbers.

4. The sum of four consecutive numbers in ar AP is 32 and the ratio of

1.	4. The sum of four consecutive numbers in ar AP is 32 and the ratio of the product of the first and the last term tothe product of two 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 = 324a = 32a = 32/4a = 8 ......(1)Now, (a - 3d)(a + 3d)/(a - d)(a + d) = 7/1515(a² - 9d²) = 7(a² - d²)15a² - 135d² = 7a² - 7d²15a² - 7a² = 135d² - 7d²8a² = 128d²Putting the value of a = 8 in above we get.8(8)² = 128d²128d² = 512d² = 512/128d² = 4d = 2So, the four consecutive numbers are 8 - (32)8 - 6 = 28 - 2 = 68 + 2 = 108 + (32)8 + 6 = 14Four consecutive numbers are 2, 6, 10 and 14