Question

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

• A. $10$
• B. $11$
• C. $12$
• D. None of these

Answer 1

The correct option is B $11$

Let the A.P be $a,a+d,a+2d,a+3d$ $a=$ first term $=11$

sum of $1^{st}$ four term = 56 $d=$ common difference

$a+a+d+a+2d+a+3d=56$

$4a+6d=56$

$4\times 11+6d=56$

$6d=12$

$6d=12$

$d=2$ Now, general term in A.P

last term = $t_n$ $t_n=a+(n-1)d$

the sum of last four terms $=112$

$t_n+t_{n-1}+t_{n-2}+t_{n-3}=112$

$\therefore a+(n-1)d+a+(n-1-1)d+a+(n-2-1)d+a+(n-3-1)d=112$

$4a+(n-1+n-2+n-3+n-4)d=112$

$4a+(4n-10)d=112$

$4\times 11+(4n-10)2=112$

$4n-10=34$

$4n=44$

$n=11$