Question

the sum of four consecutive no in an ap is 32 and the ratio of the product of the first and last term of the product of two middle term is 7 rato 15 find the number

Answer 1

Hey !!!

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Here is your answer -
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Let the number be a - 3d , a - d , a +d , a +3d

So

According to the question

a - 3d + a - d + a + d + a + 3d = 32
4a = 32
a = 8

So first term = 8

Now !!

According to the second criterion


(a - 3d ) × ( a + 3d ) / ( a - d ) ( a + d ) = 7 /15

a2 - 9d2 / a2- d2 = 7/15

so


15 ( a2 - 9d2 ) = 7 ( a2 - d2)



15a2-135d2 = 7a2-7d2

  8a2-128d2 = 0





   d2 = 4, ±2

Therefore d=4  or d=±2 1mark

So, when a=8 and d=2, the numbers are 2,6,10,14.

When a=8,d=-2 the numbers are 14,10,6,2

Answer 2

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

So,A/Q

a+3d+a-d+a+d+a+3d=32
4a=32
a=32/4
a=8 -------------(1)

Now,
(a-3d) (a+3d). 7
__________. =___
(a-d) (a+d). 15

15(a2-9d2)=7(a2-d2)
15a2-135d2=7a2-7d2
15a2-7a2=135d2-7d2
8a2=128d2

Putting the value of a from equation (1)
8(8)2=128d2
128d2=512
d2=512/128
d2=4
d=√4
d=2
So,no. Are
8-6=2
8-2=6
8+2=10
8+6=14