If the ratio of sum of m terms and n terms of an A-P- is- m 2- n 2 then prove that the ratio of its mth and nth terms will be 2 m - 1 - 2 n - 1-I know the answer for this- but i want to confirm if this method is right because everywhere- the other method is given-Sm - sn-m2 - n2On solving till a point- we get this equation2 a n-m n d-n d-2 a m-m n d-d m- 2 a n-n d-2 a m-d m- n2 a-d-m2 a-d- n-m-1In the -to prove- statement-LHS -a-m-1 d - a-n-1 dReplacing n with m using 1a-m-1 d - a-m-1 d-1 - 1-1RHS -2 m-1 - 2 n-1Replacing n with m using 1 2 m-1 - 2 m-1-1 - 1

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Common Ratio

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If the ratio of sum of m terms and n terms of an A.P. is , m²:n² then prove that the ratio of its mth and nth terms will be 2m-1 : 2n-1.

I know the answer for this, but i want to confirm if this method is right because everywhere, the other method is given!

S(m)÷s(n) = m²/n²
On solving till a point, we get this equation

2an + mnd -nd = 2am + mnd -dm
→ 2an-nd = 2am-dm
→ n(2a-d) = m(2a-d)
→ n=m. ......(1)

In the 'to prove' statement,

LHS →
a+(m-1)d / a+(n-1)d

Replacing n with m (using 1)

a+(m-1)d/a+(m-1)d = 1/1 = 1

RHS →

2m-1/2n-1
Replacing n with m (using 1)
2m-1/2m-1
= 1/1
=1

Hence, LHS=RHS.

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