The ratio of the sums of first m and n at the terms of an A.P is m^2:n^2 . Show that the ratio of it's mth and Nth terms is (2m-1) :(2n-1). Please help me guys please.
Answers
Step-by-step explanation:
We know that the sum of an arithmetic series with first term a and common difference d is S
n
=
2
n
[2a+(n−1)d]
It is given that the ratio of the sums of first m terms and first n terms of an arithmetic series is m
2
:n
2
, therefore,
S
n
S
m
=
n
2
m
2
⇒
2
n
[2a+(n−1)d]
2
m
[2a+(m−1)d]
=
n
2
m
2
⇒
n[2a+(n−1)d]
m[2a+(m−1)d]
=
n
2
m
2
⇒
[2a+(n−1)d]
[2a+(m−1)d]
=
n
2
×m
m
2
×n
⇒
[2a+(n−1)d]
[2a+(m−1)d]
=
n
m
⇒n[2a+(m−1)d]=m[2a+(n−1)d]
⇒2na+n(m−1)d=2ma+m(n−1)d
⇒2na+nmd−nd=2ma+mnd−md
⇒2na−2ma=nd−md
⇒2a(n−m)=d(n−m)
⇒2a=d
We also know that the general term of an arithmetic progression with first term a and common difference d is T
n
=a+(n−1)d.
Now using the value of d as d=2a, consider the ratio of the mth and nth terms as follows:
T
n
T
m
=
a+(n−1)d
a+(m−1)d
=
a+(n−1)(2a)
a+(m−1)(2a)
=
a+2na−2a
a+2ma−2a
=
2na−a
2ma−a
=
a(2n−1)
a(2m−1)
=
2n−1
2m−1
=(2m−1):(2n−1)
Hence, the ratio of m
th
and n
th
terms is (2m−1):(2n−1).
