Question

Divide 56 into 4 parts which are in ap such that the product of extremes to the product of mean 5:6

Answer 1

║⊕ANSWER⊕║

4 no:'s  =a-3d, a-d, a+d, a+3d

sum , 4a=56

a=14

(a-3d)(a+3d)/(a-d)(a+d)=5:6

6(a²-9d²)=5(a²-d²)

a²=54d² - 5d²

14² =49d²

d²=14² /7²

d=±14/7=±2

for d=2,

a-3d=14-6=8

a-d=14-2=12

a+d=14+2=16

a+3d=14+6=20

Numbers are 8,12,16,20

Answer 2

Answer:

Step-by-step explanation: Let the four parts be a-3d, a-d, a+d, a+3d.

Therefore, a-3d+a-d+a+d+a+3d = 56

4a = 56

a = 14

Now, a²-9d²/a²-d² = 5/6.

Put a = 14, we get, d = 2

The required terms are 8,12,16 and 20