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# The sum of first four terms of an A.P. is $56$. The sum of last four terms is $112$. If its first term is $11$ then find the number of terms.

$\begin{array}{1 1}21 \\ 20 \\ 11 \\ 19 \end{array}$

• Sum of $n$ terms of an A.P. $=S_n=\large\frac{n}{2}$$(2a+(n-1)d) Let the A.P. be a,\:a+d,\:a+2d,..........t_n Given that sum of first 4 terms of the A.P.=56 and a=11 We know that sum of n terms of an A.P. =S_n=\large\frac{n}{2}$$(2a+(n-1)d)$
$\Rightarrow\:\large\frac{4}{2}$$(2\times 11+(4-1)d)=56 \Rightarrow\:22+3d=28 \Rightarrow\:3d=6 or d=2 Step 2 Also given that sum of last 4 terms =112 Now If t_1,t_2,.........t_n is an A.P. with first term a, common difference d and last term t_n, then t_n,t_{n-1}...........t_2,t_1 is an A.P. with first term t_n and common difference -d \therefore\:t_n+t_{n-1}+t_{n-2}+t_{n-3}=112 \Rightarrow\:\large\frac{4}{2}$$\big[2.t_n+(4-1)(-2)\big]=112$
$\Rightarrow\:2t_n-6=56$
$\Rightarrow\:t_n=62/2=31$
But we know that $t_n=a+(n-1)d$
$\Rightarrow\:11+(n-1).2=31$
$\Rightarrow\:n-1=20/2=10$ $\Rightarrow\:n=11$
$i.e.,$ No. of terms $=11$.