The sum of the 5th term and the 9th term of an A.P. is 30. If its 25th

1.	The sum of the 5th term and the 9th term of an A.P. is 30. If its 25th term is three times its 18th term, find the A.P.
Answer» Solution : (Ques. Error ) $\bf{\red{\underline{\underline{\bf{Given\::}}}}}$ The sum of the 5th term and the 9th term of an A.P. is 30. If its 25th term is three times it's 8 term. $\bf{\red{\underline{\underline{\bf{To\:find\::}}}}}$ The A.P. $\bf{\red{\underline{\underline{\bf{Explanation\::}}}}}$ We know that formula of an A.P; $\boxed{\bf{a_{n}=a+(n-1)d}}}}}$ a is the term. d is the common difference. n is the term of an A.P. A/q $\longrightarrow\sf{a+(5-1)d+a+(9-1)d=30}\\\\\longrightarrow\sf{a+4d+a+8d=30}\\\\\longrightarrow\sf{2a+12d=30}\\\\\longrightarrow\sf{2(a+6d)=30}\\\\\longrightarrow\sf{a+6d=\cancel{\dfrac{30}{2} }}\\\\\longrightarrow\sf{a+6d=15.......................(1)}$ & $\longrightarrow\sf{a_{25}=3\times a_{8}}\\\\\longrightarrow\sf{a+(25-1)d=3\times [a+(8-1)d]}\\\\\longrightarrow\sf{a+24d=3\times (a+7d)}\\\\\longrightarrow\sf{a+24d=3a+21d}\\\\\longrightarrow\sf{a-3a=21d-24d}\\\\\longrightarrow\sf{\cancel{-}2a=\cancel{-}3d}\\\\\longrightarrow\sf{2a=3d}\\\\\longrightarrow\sf{a=\dfrac{3d}{2}....................(2) }$ Now; PUTTING the value of a in EQUATION (1),we GET; $\longrightarrow\sf{\dfrac{3d}{2} +6d=15}\\\\\\\longrightarrow\sf{3d+12d=30}\\\\\\\longrightarrow\sf{15d=30}\\\\\\\longrightarrow\sf{d=\cancel{\dfrac{30}{15} }}\\\\\\\longrightarrow\sf{\orange{d=2}}$ Putting the value of d in equation (2),we get; $\longrightarrow\sf{a=\dfrac{3(2)}{2} }\\\\\\\longrightarrow\sf{a=\cancel{\dfrac{6}{2}} }\\\\\\\longrightarrow\sf{\orange{a=3}}$ Thus; $\boxed{\bf{Arithmetic\:progression\::}}}$ $\bullet\:\sf{a=\boxed{3}}}\\\\\bullet\:\sf{a+d=3+2=\boxed{5}}}\\\\\bullet\:\sf{a+2d=3+2(2)=3+4=\boxed{7}}}\\\\\bullet\:\sf{a+3d=3+3(2)=3+6=\boxed{9}}}\\$

The sum of the 5th term and the 9th term of an A.P. is 30. If its 25th term is three times its 18th term, find the A.P.

Answer»

Solution : (Ques. Error )

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

The sum of the 5th term and the 9th term of an A.P. is 30. If its 25th term is three times it's 8 term.

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

The A.P.

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

We know that formula of an A.P;

$\boxed{\bf{a_{n}=a+(n-1)d}}}}}$

a is the term.
d is the common difference.
n is the term of an A.P.

A/q

$\longrightarrow\sf{a+(5-1)d+a+(9-1)d=30}\\\\\longrightarrow\sf{a+4d+a+8d=30}\\\\\longrightarrow\sf{2a+12d=30}\\\\\longrightarrow\sf{2(a+6d)=30}\\\\\longrightarrow\sf{a+6d=\cancel{\dfrac{30}{2} }}\\\\\longrightarrow\sf{a+6d=15.......................(1)}$

&

$\longrightarrow\sf{a_{25}=3\times a_{8}}\\\\\longrightarrow\sf{a+(25-1)d=3\times [a+(8-1)d]}\\\\\longrightarrow\sf{a+24d=3\times (a+7d)}\\\\\longrightarrow\sf{a+24d=3a+21d}\\\\\longrightarrow\sf{a-3a=21d-24d}\\\\\longrightarrow\sf{\cancel{-}2a=\cancel{-}3d}\\\\\longrightarrow\sf{2a=3d}\\\\\longrightarrow\sf{a=\dfrac{3d}{2}....................(2) }$

Now;

PUTTING the value of a in EQUATION (1),we GET;

$\longrightarrow\sf{\dfrac{3d}{2} +6d=15}\\\\\\\longrightarrow\sf{3d+12d=30}\\\\\\\longrightarrow\sf{15d=30}\\\\\\\longrightarrow\sf{d=\cancel{\dfrac{30}{15} }}\\\\\\\longrightarrow\sf{\orange{d=2}}$

Putting the value of d in equation (2),we get;

$\longrightarrow\sf{a=\dfrac{3(2)}{2} }\\\\\\\longrightarrow\sf{a=\cancel{\dfrac{6}{2}} }\\\\\\\longrightarrow\sf{\orange{a=3}}$

Thus;

$\boxed{\bf{Arithmetic\:progression\::}}}$

$\bullet\:\sf{a=\boxed{3}}}\\\\\bullet\:\sf{a+d=3+2=\boxed{5}}}\\\\\bullet\:\sf{a+2d=3+2(2)=3+4=\boxed{7}}}\\\\\bullet\:\sf{a+3d=3+3(2)=3+6=\boxed{9}}}\\$

The sum of the 5th term and the 9th term of an A.P. is 30. If its 25th term is three times its 18th term, find the A.P.

Solution : (Ques. Error )

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