Question

An arithmetic progression having 4 terms has sum equal to 68 and the
product of its middle two terms is 280 . Find these 4 terms

Answer 1

HLW MATE...

HERE IS YOUR CORRECT ANSWER....

THE FOUR TERMS WILL BE...

8,14,20,26..

FOR CLARIFICATION...

PLZZ REFER TO THE ABOVE ATTACHMENT...

I HOPE IT HELPS U A LOT...

TQ...FOR ASKING THIS QUESTION.....

#BE BRAINLY

Answer 2

Solution :

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

An arithmetic progression having 4 terms has sum is 68 and the products of it's middle two terms is 280.

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

These 4 terms.

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

Let the four terms of an A.P. are;

• (a-3d)
• (a-d)
• (a+d)
• (a+3d)

A/q

$\longrightarrow\sf{(a-3d)+(a-d)+(a+d)+(a+3d)=68}\\\\\longrightarrow\sf{a\cancel{-3d}+a\cancel{-d}+a\cancel{+d}+a\cancel{+3d}=68}\\\\\longrightarrow\sf{4a=68}\\\\\longrightarrow\sf{a=\cancel{\dfrac{68}{4} }}\\\\\longrightarrow\sf{\pink{a=17}}$

&

$\longrightarrow\sf{(a-d)(a+d)=280}\\\\\longrightarrow\sf{a^{2} \cancel{+ad-ad} -d^{2} =280}\\\\\longrightarrow\sf{a^{2} -d^{2} =280}\\\\\longrightarrow\sf{(17)^{2} -d^{2} =280\:\:\:\:[\therefore a=17]}\\\\\longrightarrow\sf{289-d^{2} =280}\\\\\longrightarrow\sf{-d^{2} =280-289}\\\\\longrightarrow\sf{\cancel{-}d^{2} =\cancel{-}9}\\\\\longrightarrow\sf{d=\pm\sqrt{9} }\\\\\longrightarrow\sf{\pink{d=\pm3}}$

For (+) ve common difference (d) :

$\bullet\sf{(a-3d)=17-3(3)=17-9=\boxed{8}}}\\\bullet\sf{(a-d)=17-3=\boxed{14}}}\\\bullet\sf{(a+d)=17+3=\boxed{20}}}\\\bullet\sf{(a+3d)=17+3(3)=17+9=\boxed{26}}}$

For (-) ve common difference (d) :

$\bullet\sf{(a-3d)=17-3(-3)=17-(-9)=17+9=\boxed{26}}}\\\bullet\sf{(a-d)=17-(-3)=17+3=\boxed{20}}}\\\bullet\sf{(a+d)=17+(-3)=17-3=\boxed{14}}}\\\bullet\sf{(a+3d)=17+3(-3)=17+(-9)=17-9=\boxed{8}}}$