The ratio of the sums of 1st m and 1st n terms of an AP is m^2:n^2 show that the radio of its mth and nth term is(2m-1):(2n-1) plz answer faaaast
Answers
Answer:
Answer Expert Verified ... So, the ratio of mth term and the nth term of the arithmetic series is (2m - 1):(2n -1).
Sum of m terms of an A.P. = m/2 [2a + (m -1)d]Sum of n terms of an A.P. = n/2 [2a + (n -1)d]m/2 ...
Answer:
Let S
m
and S
n
be the sum of the first m and first n terms of the AP respectively. Let, a be the first term and d be a common difference
S
n
S
m
=
n
2
m
2
⇒
2
n
[2a+(n−1)d]
2
m
[2a+(m−1)d]
=
n
2
m
2
⇒
2a+(n−1)d
2a+(m−1)d
=
n
m
⇒n[2a+(m−1)d]=m[2a+(n−1)d]
⇒2an+mnd−nd+2am+mnd−nd
⇒md−nd=2am−2an
⇒(m−n)d=2a(m−n)
⇒d=2a
Now, the ratio of mth and nth terms is
a
n
a
m
=
a+(n−1)d
a+(m−1)d
=
a+(n−1)2a
a+(m−1)2a
=
a(1+2n−2)
a(1+2m−2)
=
2n−1
2m−1
Thus, ratio of its mth and nth terms is 2m−1:2n−1