If the sum of a certain number of terms starting from first term of an A.P. is 25, 22, 19, ..., is 116. Find the last term.
Answers
Answer:
4
Step-by-step explanation:
25+22+19+16+13+10+7+4= 116
so the last term is 4
Step-by-step explanation:
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 answerANSWER
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