Question

If the ratio of sum of the first m and n terms of an A.P. is m^2 :n^2, show that the ratio of its mth and nth terms is (2m - 1): (2n-1).​

Answer 1

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

Answer 2

$\huge\blue{\boxed{\mathfrak{refer \: the \: attachment}}}$

$\small\blue{{\mathfrak{steps \: to \: do}}}$

• first write the ratio of Sum of m terms and Sum of n terms equal to m^2/n^2

• Then simplify the LHS and cut n and m in RHS

• After simplification you will find d=2a

• Then simplify whole equation and prove it easily