# The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112.

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 find the number of terms.

Solution:

Let the A.P. be $a, a+d, a+2 d, a+3 d, \ldots a+(n-2) d, a+(n-1) d$

Sum of first four terms $=a+(a+d)+(a+2 d)+(a+3 d)=4 a+6 d$

Sum of last four terms $=[a+(n-4) d]+[a+(n-3) d]+[a+(n-2) d]$

$+[a+n-1) d]$

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

According to the given condition,

$4 a+6 d=56$

$\Rightarrow 4(11)+6 d=56[$ Since $a=11$ (given) $]$

$\Rightarrow 6 d=12$

$\Rightarrow d=2$

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

$\Rightarrow 4(11)+6 d=56[$ Since $a=11$ (given) $]$

$\Rightarrow 6 d=12$

$\Rightarrow d=2$

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

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

$\Rightarrow(4 n-10) 2=68$

$\Rightarrow 4 n-10=34$

$\Rightarrow 4 n=44$

$\Rightarrow n=11$

Thus, the number of terms of the A.P. is 11.