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.
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.
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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