Question

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

• A. 12
• B. 10
• C. 11
• D. 9

Answer 1

Answer: (C)

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