Verified answer

Solution

Given :-

• Ratio of the sums of first m & n terms of an AP is = m² : n²

Prove That :-

Ratio of mth & nth terms = (2m - 1):(2n - 1)

Explanations

Let Sm & Sn be the First m & n terms of the AP respectively .

Now , Let a be the first terms & d be the common difference.

Using Formula

According to question,

==> Sm : Sn = m² : n²

==> Sm/Sn = m²/n²

==> [m{2a+(m-1)d}/2]/[n{2a+(n-1)d}/2] = m²/n²

==> [2a+(m-1)d]/[2a+(n-1)d] = m/n

==> n × [2a + (m-1)d] = m × [2a + (n-1)d]

==> 2an + nmd - nd = 2am + mnd - md

==> md - nd = 2am - 2an

==> d(m - n) = 2a( m - n)

==> d = 2a ________________(1)

We Know,

Again, According to question,

==> Ratio of mth & nth terms = [a + (m-1)d]/[a + (n-1)d]

==> Ratio of mth & nth terms = [a + (m-1)2a]/[a + (n-1)2a]

==> Radio of mth & nth terms = a(1 + 2m - 2)/a(1 + 2n - 2)

==> Ratio of mth & nth terms = (2m - 1)(2n - 1)

Hence

• Ratio of mth & nth terms will be = (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 [2a + (m -1)d] / n/2 [2a + (n -1)d] = m:n

⇒ [2a + md - d] / [2a + nd - d] = m/n

⇒ 2an + mnd - nd = 2am + mnd - md

⇒ 2an - 2am = nd - md

⇒ 2a (n -m) = d(n - m)

⇒ 2a = d

Ratio of m th term to nth term:

[a + (m - 1)d] / [a + (n - 1)d]

= [a + (m - 1)2a] / [a + (n - 1)2a]

= a [1 + 2m - 2] / a[1 + 2n -2]

= (2m - 1) / (2n -1)

So, the ratio of mth term and the nth term of the arithmetic series is (2m - 1):(2n -1)