the ratio of m term and n term of ap be m² and n² prove that n term and m term are (2m-1)(2n-1
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Step-by-step explanation:
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
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