Question

The sum of four consecutive numbers in an ap is 32 and ratio of the product of the first and the last term to the product of two 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

Hey !!

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Here is your answer
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Let the four consecutive terms of an 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-3d)(a+3d)/(a-d)(a+d)=7/15
a2-9d2/a2-d2 = 7/15
15a2-135d2= 7a2-7d2
8a2=128d2
8×(8)2 = 128d2
512/128 =d2
d2=4
d=2
So a= 8 and d =2
The numbers are
a-3d = 8-6=2
a-d=8-2=6
a+d=8+2=10
a+3d=8+6=14
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so 2 , 6 , 10 , 14

Hope it helped you !!