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NCERT Class XI Mathematics - Sequences and Series - Solutions

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Question : 26 of 106
Marks: +1, -0
The ratio of the sums of m and n terms of an A.P. is m2:n2m^2 : n^2. Show that the ratio of mth and nth term is (2m – 1) : (2n – 1).
Solution:  
Let the first term be a & common difference be d. Then
Smsn\frac{S_m}{s_n} = m2n2\frac{m^2}{n^2} ⇒ m2[2a+(m−1)d]n2[2a+(n−1)d]\frac{\frac{m}{2}[2a+(m-1)d]}{\frac{n}{2}[2a+(n-1)d]} = m2n2\frac{m^2}{n^2} ⇒ 2a+(m−1)d2a+(n−1)d\frac{2a+(m-1)d}{2a+(n-1)d} = mn\frac{m}{n}
⇒ 2an + (m – 1)nd = 2am + (n – 1)md
⇒ 2a (n – m) = (nm – m – mn + n)d ⇒ 2a = d ....(i)
Now required ratio,
a+(m−1)da+(n−1)d\frac{a+(m-1)d}{a+(n-1)d} = a+(m−1)⋅2aa+(n−1)⋅2a\frac{a+(m-1)\cdot 2a}{a+(n-1)\cdot 2a} [by (i)]
= 1+2m−21+2n−2\frac{1+2m-2}{1+2n-2} = 2m−12n−1\frac{2m-1}{2n-1}
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