Question

the sum of 4 consecutive numbers in an AP is 32 and the ratio of the product of the first and the last term to the product of the 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
Answer

Answer 2

Step-by-step explanation:

Let the 4 consecutive numbers in AP be

(a−3d),(a−d),(a+d),(a+3d)

According to the question,

a−3d+a−d+a+d+a+3d=32

4a=32

a=8

Now,

(a−d)(a+d)(a−3d)(a+3d)=157

15(a2−9d2)=7(a2−d2)

15a2−135d2=7a2−7d2

8a2=128d2

Putting the value of a, we get,

d2=4

d=±2

So, the four consecutive numbers are 2,6,10,14 or 14,10,6,2.