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Q.
The sum of the first four terms of an A.P. is $56$ . The sum of the last four terms is $112$ . If its first term is $11$ , then the number of terms is
Sequences and Series
Solution:
Let the A.P. is $a, a+d, a+2 d, \ldots \ldots$
Given, sum of first four terms $=56$
i.e., $ T_1+T_2+T_3+T_4=56$
$\Rightarrow a+a+d+a+2 d+a+3 d=56$
$\Rightarrow 4 a+6 d=56$
$\Rightarrow 4 \times 11+6 d=56 (\because a=11)$
$\Rightarrow 6 d=56-44=12$
$\Rightarrow d=\frac{12}{6}$
$\Rightarrow d=2$
If last term is $T_n$, then sum of last four terms
$ =T_n+T_{n-1}+T_{n-2}+T_{n-3}=112 $
$ \therefore a+(n-1) d+a+(n-1-1) d+a+(n-2-1) d $
$+a+(n-3-1) d=112\left[\because T_n=a+(n-1) d\right]$
$ \Rightarrow 4 a+d(n-1+n-2+n-3+n-4)=112 $
$ \Rightarrow 4 \times 11+2[4 n-10]=112 (\because a=11, d=2)$
$ \Rightarrow 44+8 n-20=112 $
$ \rightarrow 8 n-112-44+20 $
$ \Rightarrow 8 n=132-44 $
$ \Rightarrow 8 n=88$
$ \Rightarrow n=\frac{88}{8} $
$ =11 $
Hence, total numbers of terms in the sequence $=11$