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.

Answer 1

Let the A.P. be $a,a+d,a+2d,a+3d,...a+(n-2)d,a+(n-1)d.$
Sum of first four terms $=a+(a+d)+(a+2d)+(a+3d)=4a+6d$
Sum of last four terms
$=[a+(n-4\left)d\right]+[a+(n-3\left)d\right]+[a+(n-2\left)d\right]+[a+(n-1\left)d\right]\Rightarrow =4a+(4n-10)d$
According to the given condition, $4a+6d=56$
$\Rightarrow 4\left(11\right)+6d=56[\text{Since}a=11\text{(given)}]\Rightarrow 6d=12\Rightarrow d=2\therefore 4a+(4n-10)d=112\Rightarrow 4\left(11\right)+(4n-10)2=112\Rightarrow (4n-10)2=68\Rightarrow 4n-10=34\Rightarrow 4n=44\Rightarrow n=11$
Thus the number of terms of A.P. is $11.$