Question

The sum of four consecutive number in AP is 32 and the ratio of the product of the first and the last term of the product of two middle term is 7:15 .find the sum.

Answer 1

Hey !!

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Here is your answer

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

So,
        a-3d+a-d+a+d+a+3d=32

       4a=32

      a=8

      (a-3d)(a+3d) / (a-d)(a+d)=7/15

      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

Answer:

 2 , 6, 10 , 14 .

Step-by-step explanation:

Let four consecutive number as

a - 3 d , a - d , a + d , a + 3 d .

Given their sum is 32 .

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

4 a = 32

a = 8 .

Now :

Also given the ratio of the product of the first and the last term to product of two middle term to the product of two middle terms is 7:15 .

$\displaystyle{\dfrac{(a-3d)(a+3d)}{(a-d)(a+d)} = \dfrac{7}{15} }\\\\\\\displaystyle{\dfrac{a^2-(3d)^2}{a^2-d^2} = \dfrac{7}{15} }$

We have a = 8

$\displaystyle{\dfrac{64-(3d)^2}{64-d^2} = \dfrac{7}{15} }\\\\\\\displaystyle{\dfrac{64-9d^2}{64-d^2} = \dfrac{7}{15} }$

960 - 135 d² = 448 - 7 d²

135 d² - 7 d²  = 960 - 448

128 d²  = 512

d²  = 4

d = ± 2

When d = 2

Numbers are ,

a - 3 d , a - d ,  a + d , a + 2 d

= > 8 - 6 = 2

= > 8 - 2 = 6

= > 8 + 2 = 10

= > 8 + 6 = 14

When d = - 2

= > 8 + 6 = 14

= > 8 + 2 = 10

= > 8 - 2 = 6

= > 8 - 6 = 2 .

But both ways we get same numbers.

i.e.  2 , 6, 10 , 14 .