Question

The sum of
first three terms in an AP is 4
to the sum of the squares is 224. Find the
find three terms of this A.P.​

Answer 1

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Let the terms be a-d, a, a+d

sum=24

a-d+a+a+d=3a=24

a=8

(a-d) ^2+a^2+(a+d)^2=224

a^2+2(a^2+d^2)=224

3a^2+2d^2=224

3(64)+2d^2=224

2d^2=224-192

d^2=32/2

d=√16=4 put in numbers(assumed )

check calculation.. procedure will b same..

hope these will help you....

please mark as brainliest....

Answer 2

Solution :(Question Error)

$\bf{\red{\underline{\bf{Given\::}}}}$

The sum of first three terms in an A.P. is 24 to the sum of the square is 224.

$\bf{\red{\underline{\bf{To\:find\::}}}}$

The three terms of this A.P.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

let the three terms be a-d, a, a+d

A/q

$\longrightarrow\sf{a\cancel{-d}+a+a\cancel{+d}=24}\\\\\longrightarrow\sf{3a=24}\\\\\longrightarrow\sf{a=\cancel{\dfrac{24}{3} }}\\\\\longrightarrow\sf{\pink{a=8}}$

Now;

The sum of the square is 224 :

$\longrightarrow\sf{(a-d)^{2} +(a)^{2} +(a+d)^{2} =224}\\\\\longrightarrow\sf{a^{2} +d^{2} \cancel{-2ad}+a^{2} +a^{2} +d^{2} \cancel{+2ad}=224}\\\\\longrightarrow\sf{3a^{2} +2d^{2} =224}\\\\\longrightarrow\sf{3(8)^{2} +2d^{2} =224\:\:\:\:\:[\therefore a=8]}\\\\\longrightarrow\sf{3*64+2d^{2} =224}\\\\\longrightarrow\sf{192+2d^{2} =224}\\\\\longrightarrow\sf{2d^{2} =224-192}\\\\\longrightarrow\sf{2d^{2} =32}\\\\\longrightarrow\sf{d^{2} =\cancel{\dfrac{32}{2} }}$

$\longrightarrow\sf{d^{2} =16}\\\\\longrightarrow\sf{d=\sqrt{16} }\\\\\longrightarrow\sf{\pink{d=4}}$

Thus;

$\bullet\sf{The\:First\:term,(a-d)=(8-4)=4}\\\bullet\sf{The\:Second\:term,(a)=8}\\\bullet\sf{The\:Third\:term,(a+d)=(8+4)=12}$