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

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.