Question

The sum of four consecutive
numbers in an AP is 32 and the ratio of the product of the first
and last term to the product of two middle term is 7:15. Find the numbers.
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Answer 1

Answer:

Let the four consecutive numbers in AP be (a−3 d),(a−d),(a+d) and (a+3 d)

So, according to the question.

a−3 d+a−d+a+d+a+3 d=32

4 a=32

a=32/4

a=8     (1)

Now, (a−3 d)(a+3 d)/(a−d)(a+d)=7/15

15(a²−9 d²)=7(a²−d²)

15 a²−135 d²=7 a²−7 d²

15 a²−7 a²=135 d²−7 d²  

8 a²=128 d²

Putting the value of a=8 in above we get.

8(8)²=128 d²

128 d²=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