Question

the sum of four consecutive numbers in an APIs 32 and the ratio of the product of two middle term is 7:15.find the numbers

Answer 1

Hello friend

Let the terms be

a-3d , a-d , a+d , a+3d

Their sum

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

4a = 32

a = 8

then ATQ

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

(a)^2 - (3d)^2 / (a)^2 - (d)^2 = 7/15

(8)^2 - (3d)^2 / (8)^2 - (d)^2 = 7/15

64 - 9d^2 / 64 - d^2 = 7/15

15(64 - 9d^2) = 7(64 - d^2)

960 - 135d^2 = 448 - 7d^2

960 - 448 = -7d^2 + 135d^2

512 = 128d^2

d^2 = 4

d = 2

so terms are

a-3d = 8 - 3(2) = 8 - 6 = 2

a-d = 8 - 2 = 6

a = 8

a + d = 8 + 2 = 10

a + 3d = 8 + 3(2) = 8 + 6 = 14


Hope it helps u . Have a nyc day .

Answer 2

Hi..... Hope it helped