Question

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Find 4 terms in an AP whose sum is 72 and the ratio of the product of the 1st and 4th terms to the product of the 2nd and 3rd term is 9:10.

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Answer 1

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Answer 2

$\large{\underline{\underline{\mathfrak{\blue{\sf{Correct\:question-}}}}}}$

Find 4 consecutive terms in an AP whose sum is 72 and the ratio of the product of the 1st and 4th terms to the product of the 2nd and 3rd term is 9:10.

$\large{\underline{\underline{\mathfrak{\pink{\sf{Answer-}}}}}}$

The four consecutive terms be : $\sf{12,16,20\:and\:24}$

$\large{\underline{\underline{\mathfrak{\blue{\sf{Explanation-}}}}}}$

Let's suppose the four consecutive terms be : $\sf{a,a+d,a+2d\:and\:a+3d}$

It is given that sum of these terms is 72"

$\implies$ $\sf{(a)+(a+d)+(a+2d)+(a+3d)=72}$

$\implies$ $\sf{a+a+d+a+2d+a+3d=72}$

$\implies$ $\sf{4a+6d=72}$

By taking 2 as common,

$\implies$ $\red{\sf{2a+3d=36}}$ _____(1)

___________________

Product of 1st and 4th term = $\bold{\sf{a(a+3d)}}$

Product of 2nd and 3rd term = $\bold{\sf{(a+d)(a+2d)}}$

It is also given that the ratio of the product of the 1st and 4th terms to the product of the 2nd and 3rd term is 9:10"

$\sf{a(a+3d):(a+d)(a+2d)}$ = $\sf{9:10}$

$\implies$ ${\sf\dfrac{(a)(a + 3d)}{(a + d)(a + 2d)} = \dfrac{9}{10}}$

By cross multiplying,

$\implies$ $\sf{10(a^2+3ad)=9(a^2+2ad+ad+2d^2)}$

$\implies$ $\sf{10a^2+30ad=9a^2+27ad+18d^2}$

$\implies$ $\sf{10a^2-9a^2+30ad-27ad=18d^2}$

$\implies$ $\sf{a^2+3ad=18d^2}$

$\implies$ $\sf{a^2+3ad-18d^2=0}$

By making factors,

$\implies$ $\sf{(a+6d)(a-3d)=0}$

$\implies$ $\sf{(a+6d)=0\:and\:(a-3d)=0}$

$\implies$ $\red{\sf{a=-6d\:and\:3d}}$

So, there are two cases.

$\bold\green{Case\:1-}$

$\large{\underline{\boxed{\sf{\orange{When\:a=-6d}}}}}$

put a = -6d in equation 1.

$\sf{2(-6d)+3d=36}$

$\implies$ $\sf{-12d+3d=36}$

$\implies$ $\sf{-9d=36}$

$\implies$ $\sf{d=\cancel\dfrac{36}{-9}}$

$\implies$ $\large\red{\sf{d=-4}}$

Now put d = -4 in a = -6d.

$\implies$ $\sf{a=-6(-4)}$

$\implies$ $\large\red{\sf{a=24}}$

So, consecutive terms in case 1 are :

★ 1st term : 24

$\sf{a=24}$

★ 2nd term : 20

$\sf{a+d=24+(-4)}$

$\implies$ $\sf{24-4=20}$

★ 3rd term : 16

$\sf{a+2d=24+2(-4)}$

$\implies$ $\sf{24-8=16}$

★ 4th term : 12

$\sf{a+3d=24+3(-4)}$

$\implies$ $\sf{24-12=12}$

_________________________

$\bold\green{Case\:2-}$

$\large{\underline{\boxed{\sf{\orange{When\:a=3d}}}}}$

put a = 3d in equation 1.

$\sf{2(3d)+3d=36}$

$\implies$ $\sf{6d+3d=36}$

$\implies$ $\sf{9d=36}$

$\implies$ $\sf{d=\cancel\dfrac{36}{9}}$

$\implies$ $\large\red{\sf{d=4}}$

Now put d = 4 in a = 3d.

$\implies$ $\sf{a=3(4)}$

$\implies$ $\large\red{\sf{a=12}}$

So, consecutive terms in case 1 are :

★ 1st term : 12

$\sf{a=12}$

★ 2nd term : 16

$\sf{a+d=12+4}$

$\implies$ $\sf{12+4=16}$

★ 3rd term : 20

$\sf{a+2d=12+2(4)}$

$\implies$ $\sf{12+8=20}$

★ 4th term : 24

$\sf{a+3d=12+3(4)}$

$\implies$ $\sf{12+12=24}$

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