Question

Sum of 4 consecutive numbers in an AP is 32 and the ratio of the first and last term to the product of two middle terms is 7:15. Find the numbers

Answer 1

Solution:-

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

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Answer 2

Answer:

Let the four consecutive term of the AP be ,

a -  3 d , a - d , a + d , a + 3 d  

A.T.Q.

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

4 a = 32

a = 8

and  

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

15  ( a²  - 9 d² ) = 7 ( a²  -  d² )

8 a²  =  128 d² [ a = 8 ]  

d = ± 2

Therefore , numbers are 2 , 6 , 10 , 14 or 14 , 10 , 6 , 2 .