If the nth term of an AP
is 1/n and nth term is 1/m
show that the
sum of mn terms is 1/2(MN+1)
Answers
Answered by
2
Answer:
m
=
n
1
=a+(m−1)d ......(1)
a
n
=
m
1
=a+(n−1)d .......(2)
Subtracting equation 2 form equation 1, we get,
(m−n)d=
n
1
−
m
1
(m−n)d=
mn
m−n
d=
mn
1
Putting this value of d in equation 1, we get,
a+(m−1)
mn
1
=
n
1
a+
n
1
−
mn
1
=
n
1
a=
mn
1
S
mn
=
2
mn
[2a+(mn−1)d]
S
mn
=
2
mn
[
mn
2
+(mn−1)×
mn
1
]
S
mn
=
2
1
(mn+1)
Answered by
1
Answer:
m
=
n
1
=a+(m−1)d ......(1)
a
n
=
m
1
=a+(n−1)d .......(2)
Subtracting equation 2 form equation 1, we get,
(m−n)d=
n
1
−
m
1
(m−n)d=
mn
m−n
d=
mn
1
Putting this value of d in equation 1, we get,
a+(m−1)
mn
1
=
n
1
a+
n
1
−
mn
1
=
n
1
a=
mn
1
S
mn
=
2
mn
[2a+(mn−1)d]
S
mn
=
2
mn
[
mn
2
+(mn−1)×
mn
1
]
S
mn
=
2
1
(mn+1)
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