The sum of first 4 terms of an A.P. is 56. The sum of last 4 terms is

1.	The sum of first 4 terms of an A.P. is 56. The sum of last 4 terms is 112 if it's 1st term is 11 then find the no. Of terms
Answer» SOLUTION : $\underline{\bf{Given\::}}}}$ Sum of first 4 TERMS of an A.P. = 56 The sum of LAST 4 term = 112 First term (a) = 11 $\underline{\bf{Explanation\::}}}}$ Let the four term's of an A.P are a, a+d , a+2d , a+3d $\underline{\boldsymbol{According\:to\:the\:question\::}}}$ $\longrightarrow\sf{a+a+d+a+2d+a+3d=56}\\\\\longrightarrow\sf{4a+6d=56}\\\\\longrightarrow\sf{4(11)+6d=56\:\:[\therefore a=11]}\\\\\longrightarrow\sf{44+6d=56}\\\\\longrightarrow\sf{6d=56-44}\\\\\longrightarrow\sf{6d=12}\\\\\longrightarrow\sf{d=\cancel{12/6}}\\\\\longrightarrow\bf{d=2}$ ∴ We know that formula of the nth term : $\boxed{\bf{a_n=a+(n-1)d}}}$ a+(n-1)d , a+(n-2)d , a+(n-3)d , a+(n-4)d $\longrightarrow\sf{a+(n-1)d+a+(n-2)d+a+(n-3)d+a+(n-4)d=112}\\\\\longrightarrow\sf{4a+(4n-10)d=112}\\\\\longrightarrow\sf{4(11)+(4n-10)(2)=112}\\\\\longrightarrow\sf{44+8n-20=112}\\\\\longrightarrow\sf{8n+24=112}\\\\\longrightarrow\sf{8n=112-24}\\\\\longrightarrow\sf{8n=88}\\\\\longrightarrow\sf{n=\cancel{88/8}}\\\\\longrightarrow\bf{n=11}$ Thus; The number of terms of an A.P. will be 11 .

The sum of first 4 terms of an A.P. is 56. The sum of last 4 terms is 112 if it's 1st term is 11 then find the no. Of terms

Answer»

SOLUTION :

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

Sum of first 4 TERMS of an A.P. = 56
The sum of LAST 4 term = 112
First term (a) = 11

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

Let the four term's of an A.P are a, a+d , a+2d , a+3d

$\underline{\boldsymbol{According\:to\:the\:question\::}}}$

$\longrightarrow\sf{a+a+d+a+2d+a+3d=56}\\\\\longrightarrow\sf{4a+6d=56}\\\\\longrightarrow\sf{4(11)+6d=56\:\:[\therefore a=11]}\\\\\longrightarrow\sf{44+6d=56}\\\\\longrightarrow\sf{6d=56-44}\\\\\longrightarrow\sf{6d=12}\\\\\longrightarrow\sf{d=\cancel{12/6}}\\\\\longrightarrow\bf{d=2}$

∴ We know that formula of the nth term :

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

a+(n-1)d , a+(n-2)d , a+(n-3)d , a+(n-4)d

$\longrightarrow\sf{a+(n-1)d+a+(n-2)d+a+(n-3)d+a+(n-4)d=112}\\\\\longrightarrow\sf{4a+(4n-10)d=112}\\\\\longrightarrow\sf{4(11)+(4n-10)(2)=112}\\\\\longrightarrow\sf{44+8n-20=112}\\\\\longrightarrow\sf{8n+24=112}\\\\\longrightarrow\sf{8n=112-24}\\\\\longrightarrow\sf{8n=88}\\\\\longrightarrow\sf{n=\cancel{88/8}}\\\\\longrightarrow\bf{n=11}$

The sum of first 4 terms of an A.P. is 56. The sum of last 4 terms is 112 if it's 1st term is 11 then find the no. Of terms

SOLUTION :

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