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The ratio of sum of first …

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The ratio of sum of first m and n terms of an AP is m^2:n^2 . Show that the ratio of mth term and nth term is (2m-1):(2n-1).

Posted by Divanshu Rawat 6 years, 10 months ago

CBSE > Class 10 > Mathematics

3 answers

Aditya Singh 6 years, 10 months ago

Sm:Sn=m^2:n^2 Sm/Sn=m/2(2a+(n-1)d)/n/2(2a+(n-1)d ) Sm/Sn=(2a+(m-1)d)/(2a+(n-1)d) =m^2/n^2×n/m=m/n 2a+(m-1)d/2a+(n-1)d=m/n By cross multiplication 2an+n(m-1)d=2am+m(n-1)d 2an+nmd-nd=2am+mnd-md 2an-nd=2am-md 2an-2am=nd-md 2a(n-m)=d(n-m) 2a=d am/an=a+(m-1)d/a+(n-1)d am/an=a+(m-1)2a/a+(n-1)2a am/an=a+2am-2a/a+2an-2a am/an=2am-a/2an-a am/an=a(2m-1)/a(2n-1)=2m-1/2n-1

Gaurav Seth 6 years, 10 months ago

Let S_m and S_n be the sum of the first m and first n terms of the A.P. respectively. Let, a be the first terms and d be a common difference.

⇒[2a+(m−1)d]
n =[2a(n−1)d]m
⇒2an+mnd−nd=2am+mnd−md
⇒md−nd=2am−2an
⇒(m−n)d=2a(m−n)
⇒d=2a
Now, the ratio of its mth and nth terms is

Thus, the ratio of its mth and nth terms is 2m – 1 : 2n – 1.

Divanshu Rawat 6 years, 10 months ago

Freinds please answer it

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