Find the last term of the AP 25,22,19.........? whose sum is 101
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Answer:
we have,
A.P. is
25,22,19......
S
n
=116
a=25
d=T
2
−T
1
d=22−25
d=−3
Then, we know that
S
n
=
2
n
(2a+(n−1)d)
⇒116=
2
n
(2×25+(n−1)×(−3))
⇒116=
2
n
(50−3n+3)
⇒116=
2
n
(53−3n)
⇒232=n(53−3n)
⇒3n
2
−53n+232=0
⇒3n
2
−(29+24)n+232=0
⇒3n
2
−29n−24n+232=0
⇒n(3n−29)−8(3n−29)=0
⇒(3n−29)(n−8)=0
If
3n−8=0
n=
3
8
(notpossible)
If
n−8=0
n=8
So,
The last term is
S
n
=
2
n
(a+l)
116=
2
8
(25+l)
116=4(25+l)
116=100+4l
116−100=4l
4l=16
l=4
Hence, the last term of this series is
4.
This is the answer.
Step-by-step explanation:Easy Explained please mark me as Brainlist
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