Question

the sum a pf the first three term of an ap is 33 if the product of first term and thirs etrm exceeds the 2nd term by 29 then find the ap​

Answer 1

Answer:

A.P. = 20 , 11 , 2.....

A.P. = 2 , 11 , 20....

Step-by-step explanation:

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SEE THE ATTACHMENT MY FRIEND....

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Answer 2

Solution :

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

The sum of the the first three term of an A.P. is 33, If the product of first term and third term exceeds the 2nd term by 29.

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

The arithmetic progression.

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

Let the first three terms be ;

• (a-d)
• a
• (a+d)

A/q

$\longrightarrow\sf{(a-d)+a+(a+d)=33}\\\\\longrightarrow\sf{a\cancel{-d}+a+a \cancel{+d}=33}\\\\\longrightarrow\sf{3a=33}\\\\\longrightarrow\sf{a=\cancel{\dfrac{33}{3} }}\\\\\longrightarrow\sf{\blue{a=11}}$

&

$\longrightarrow\sf{(a-d)(a+d)=a+29}\\\\\longrightarrow\sf{a^{2} \cancel{+ad-ad}-d^{2} =a+29}\\\\\longrightarrow\sf{a^{2} -d^{2} =a+29}\\\\\longrightarrow\sf{(11)^{2} -d^{2} =11+29\:\:\:[\therefore a=11]}\\\\\longrightarrow\sf{121-d^{2} =40}\\\\\longrightarrow\sf{-d^{2} =40-121}\\\\\longrightarrow\sf{\cancel{-}d^{2} =\cancel{-}81}\\\\\longrightarrow\sf{d^{2} =81}\\\\\longrightarrow\sf{d=\pm\sqrt{81} }\\\\\longrightarrow\sf{\blue{d=\pm9}}$

Thus;

$\dag\:\boxed{\bf{\overline{\mid{Arithmetic\:progression\mid}}}}}$

For + ve common difference use :

$\bullet{\sf{(a-d)=(11-9)=\boxed{2}}}}\\\bullet{\sf{(a)=\boxed{11}}}}\\\bullet{\sf{(a+d)=(11+9)=\boxed{20}}}}$

For - ve common difference use :

$\bullet{\sf{(a-d)=[11-(-9)]=[11+9]=\boxed{20}}}}\\\bullet{\sf{(a)=\boxed{11}}}}\\\bullet{\sf{(a+d)=[11+(-9)]=[11-9]=\boxed{2}}}}$