Question

10) The sum of 4 consecutive numbers in an AP is
32 & the ratio of the product of the first and the
last terms to the product of the two middle terms
is 7:15. Find the numbers​

Answer 1

Step-by-step explanation:

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

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 2,6,10and14

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