If the sum of a certain number of terms starting from first term

Solution:

In the given problem, we have the sum of the certain number of terms of an A.P. and we need to find the last term for that A.P.

So here, let us first find the number of terms whose sum is 116. For that, we will use the formula,

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So for the given A.P $(25,22,19, \ldots)$

The first term (a) = 25

The sum of $n$ terms $S_{n}=116$

Common difference of the A.P. $(d)=a_{2}-a_{1}$

$=22-25$

 

$=-3$

So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

$116=\frac{n}{2}[2(25)+(n-1)(-3)]$

$116=\left(\frac{n}{2}\right)[50+(-3 n+3)]$

$116=\left(\frac{n}{2}\right)[53-3 n]$

$(116)(2)=53 n-3 n^{2}$

So, we get the following quadratic equation,

$3 n^{2}-53 n+232=0$

On solving by splitting the middle term, we get,

$3 n^{2}-24 n-29 n+232=0$

$3 n(n-8)-29(n-8)=0$

 

$(3 n-29)(n-8)=0$

Further,

$3 n-29=0$

$n=\frac{29}{3}$

Also,

$n-8=0$

 

$n=8$

Now, since n cannot be a fraction, so the number of terms is 8.

So, the term is a8

$a_{2}=a_{1}+7 d$

$=25+7(-3)$

$=25-21$

 

$=4$

Therefore, the last term of the given A.P. such that the sum of the terms is 116 is.