If the mth term of an AP is 1/n and nth term is 1/m then show that sum of mn term is 1/2(mn +1)
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Answers
Answered by
3
Answer:
a
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)
Step-by-step explanation:
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Answered by
1
Answer:
==n1=a+(m−1)d ......(1)
an=m1=a+(n−1)d .......(2)
Subtracting equation 2 form equation 1, we get,
(m−n)d=n1−m1
(m−n)d=mnm−n
d=mn1
Putting this value of d in equation 1, we get,
a+(m−1)mn1=n1
a+n1−mn1=n1
a=mn1
Smn=2mn[2a+(mn−1)d]
Smn=2mn[mn
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