If the sum of a certain number of terms of the A.P. 25,22,19,... is 116. Find the last term.
Let a be the first term and d be the common difference of the given AP. Then a=25,d=−3.
Let sum of first n terms of the given AP is 116. Therefore,
Sn=116⇒n2[2a+(n−1)d]=116
⇒n2[2×25+(n−1)×−3]=116
⇒n[53−3n]=232
⇒3n2−53n+232=0
⇒3n2−24n−29n+232=0
⇒(3n−29)(n−8)=0
⇒n=293,8
But, n=293 is not possible.
∴n=8
Now, last term =a8=a+7d=25+7×−3=4