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sum of the term of an A.P 25,22,19 _ _ 116 is find the value of n and also find 15th term of an A.P ​

Answer»

we have,A.P. is25,22,19......Sn=116a=25d=T2−T1d=22−25d=−3Then, we KNOW thatSn=2n(2a+(n−1)d)⇒116=2n(2×25+(n−1)×(−3))⇒116=2n(50−3n+3)⇒116=2n(53−3n)⇒232=n(53−3n)⇒3n2−53n+232=0⇒3n2−(29+24)n+232=0⇒3n2−29n−24n+232=0⇒n(3n−29)−8(3n−29)=0⇒(3n−29)(n−8)=0If3n−8=0n=38(notpossible)Ifn−8=0n=8So,The last TERM isSn=2n(a+l)116=28(25+l)116=4(25+l)116=100+4l116−100=4l4l=16l=4Hence, the last term of this series is4.This is the ANSWER.


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