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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.

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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) ]

$=6 d=12$

$=d=2$

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

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

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

$\Rightarrow 4 n-10=34$

$\Rightarrow 4 n=44$

$\Rightarrow n=11$

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

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