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# Question-The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and last term to the product of two middle term is 7 : 15.find numbers.

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### Question-The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and last term to the product of two middle term is 7 : 15.find numbers.

SOLUTION-
Let the four consecutive terms of an AP are a – 3d, a – d, a + d, a  + 3d

According to question that the sum of these term is 32

Therefore a – 3d + a – d + a + d + a  + 3d  = 32

4a = 32d

a = 8

${\color{DarkBlue} \boldsymbol{\frac{\left ( a-3d \right )\left ( a+3d \right )}{\left ( a-d \right )\left ( a+d \right )}=\frac{7}{15}}}$

${\color{DarkBlue} \boldsymbol{\frac{a^{2}-9d^{2}}{a^{2}-d^{2}}=\frac{7}{15}}}$

15a² – 135d² = 7a² – 7d²

15a² – 7a² = 135d² – 7d²

8a² = 128d²

a² = 16d²

Putting the value of a = 8

16d² = 8² = 64

${\color{DarkBlue} \boldsymbol{d^{2}=\frac{64}{16}=4}}$

${\color{DarkBlue} \boldsymbol{d =\pm 2}}$

If d = 2 then the terms will be

a – 3d = 8 – 3×2 = 2, a – d = 8 – 2 =6 ,a + d = 8 + 2= 10 ,a  + 3d = 8 + 3×2 =14

If d = –2 then a – 3d = 8 – 3 × –2 = 14, a – d  = 8 – (–2) = 10, a + d = 8 + (–2) = 6, a  + 3d = 8 + 3× –2= 2

Therefore the terms  of AP  are 2,6,10 and 14 or 14, 10, 6 and 2

Alternative Method-

Let the four terms of AP are a, a + d, a + 2d, a + 3d

As it is given that sum of these terms is 32

Therefore a + a + d + a + 2d + a + 3d = 32

4a + 6d = 32

2a + 3d  = 16………….(i)

We are given that the ratio between the product of first and last term is 7 : 15

${\color{DarkBlue} \boldsymbol{\frac{a\left ( a+3d \right )}{\left ( a+d \right )\left ( a+2d \right )}=\frac{7}{15}}}$

15a² + 45ad = 7a² + 14ad  + 7ad + 14d²

8a²  + 24ad  – 14d² = 0

4a² + 12ad – 7d² = 0

Substituting 2a + 3d = 16⇒ a = (16 – 3d)/2  in equation (ii)

${\color{DarkBlue} \boldsymbol{4\frac{\left ( 16-3d \right )^{2}}{4}+12d\frac{\left ( 16-3d \right )}{2}-7d^{2}=0}}$

256 + 9d² – 96d + 96d – 18d² – 7d² = 0

–16d²  + 256 = 0

16d² = 256

d² = 16

d = ±4

Substituting value of d = 4  in equation number (i)  2a + 3d = 16 value of a = 2  and if d =–4 then a = 14

When d = 4

First term  a = 2, second term a + d = 2 + 4 = 6, third term = a + 2d = 2 + 2 ×4 = 10, fourth term = a + 3d = 2 + 3 ×4 = 14

When d = –4

First term = 14, second term a +d= 14 –4 = 10, third term  a +2d =14 –2×4 =6, fourth term a + 3d = 14 –3×4 = 2

Therefore the reqired term of AP 2,6,10,14 or 14,10,6,2

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