Question

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

Answer 1

Answer:

Four consecutive numbers are 2, 6, 10 and 14

Step-by-step explanation:

Lets consider four consecutive numbers in AP be (a - 3d), (a - d), (a + d) and (a + 3d)

So, according to the given data the sum of four consecutive numbers is 32

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

4a = 32

a = 32/4

a = 8 ......(1)

Now, ratio of the product of the  first and the last terms to the product of the two middle terms in 7:15 i,e

(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 required 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

Answer 2

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