Sum of first m terms of an AP is 0. If the first term
is given by a, then the sum of next n terms is
an(m,n)
(1)
2an(m,n)
m+1
(2)
1-m
an(m + n)
(3)
an(m+n)
1-m
(4)
1+m
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Answer:
Step-by-step explanation:
ANSWER
Let a be the first term and d be the common difference of the given AP. Then,
S
n
=
2
n
[2a+(n−1)d]
Given,
S
m
=n
2
m
[2a+(m−1)d]=n
2am+m(m−1)d=2n ...........(1)
S
n
=m
2
n
[2a+(n−1)d]=m
2an+n(n−1)d=2m ..........(2)
On subtracting 2 from 1, we get,
2a(m−n)+[(m
2
−n
2
)−(m−n)]d=2(n−m)
(m−n)[2a+(m+n−1)d]=2(n−m)
2a+(m+n−1)d=−2 ..........(3)
Sum of (m+n) terms of the given AP
S
m+n
=
2
m+n
[2a+(m+n−1)d]
=
2
m+n
(−2)=−(m+n)
Hence proved.
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