Question

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

Answer 1

Answer:

Step-by-step explanation:

Let the 4 consecutive terms in AP be (a-3d) , (a-d) , (a+d) and (a+3d)

The sum of the 4 consecutive numbers is,

(a-3d) + (a-d) + (a+d) + (a+3d)=32

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

⇒4a=32⇒a=8.  

Product of first and last term= (a-3d)(a+3d)=a²-(3d)²=8²-9d²=64-9d²

Product of two middle terms=(a-d)(a+d)=a²-d²=8²-d²=64-d²

Product of first and last terms/Product middle terms=7/15

(64-9d²)/(64-d²)=7/15

cross multiply,

7(64-d²)=15(64-9d²)

448-7d²=960-135d²

135d²-7d²=960-448

128d²=512

d²=512/128=4⇒d²=4⇒d=±√4⇒d=±2

CASE-1:

a=8 and d=2

The numbers are:

2,6,10,14

CASE-2:

a=8 and d= -2

The numbers are,

14,10,6,2.

Hence found the numbers.