Question

2.) The ratio of the sums of m and n terms of an A.P. is m?: n². Show that the ratio
of mth and nth term is (2m - 1): (2n - 1).
​

Answer 1

Answer:

2.) The ratio of the sums of m and n terms of an A.P. is m?: n². Show that the ratio

of mth and nth term is (2m - 1): (2n - 1).

2.) The ratio of the sums of m and n terms of an A.P. is m?: n². Show that the ratio

of mth and nth term is (2m - 1): (2n - 1).

Answer 2

Answer:

2m−1:2n−1

Step-by-step explanation:

Let Smand Sn

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 differenceSnS m = n2 m2

⇒

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

Hopes it helps

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