Question

the sum of four consecutive terms which are in arithnetic progression is 32 and ratio of the product first and last term to the product of two middle terms in 7:5 find the number​

Answer 1

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 - (3*2)

8 - 6 = 2

8 - 2 = 6

8 + 2 = 10

8 + (3*2)

8 + 6 = 14

Four consecutive numbers are 2, 6, 10 and 14

Cheers. Mark brainliest. Sub to pewds.