Question

If the ratio of sum of the first m and n terms of an A.P. is m2 : n2 , show that

the ratio of its mth and nth terms is (2m -1): (2n -1).​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Let $a_n$ and $d$ be the nth term and common difference of the A.P respectively

Then, $a_n = a_1 + (n-1)d$

Here,

Sum of first m terms of the A.P

$=\sum\limits^{m}_{k=1}a_k=\frac{m}{2} \left[2a_1+(m-1)d\right]$

Sum of first n terms of the A.P

$=\sum\limits^{n}_{k=1}a_k=\frac{n}{2} \left[2a_1+(n-1)d\right]$

Given,

$\frac{ \sum\limits^{m}_{k=1}a_k}{ \sum\limits^{n}_{k=1}a_k}=\frac{m^2}{n^2}\\\\\implies \frac{\frac{m}{2} \left[2a_1+(m-1)d\right]}{\frac{m}{2} \left[2a_1+(n-1)d\right]}=\frac{m^2}{n^2}\\\implies\frac{m}{n} \times\frac{2a_1+(m-1)d}{2a_1+(n-1)d} =\frac{m^2}{n^2}\\\implies\frac{2a_1+(m-1)d}{2a_1+(n-1)d}=\frac{m}{n}\\\implies n\left[2a_1+(m-1)d\right]=m\left[2a_1+(n-1)d\right]\\\implies 2a_1n+(m-1)dn=2a_1m+(n-1)dm\\\implies (m-1)dn-(n-1)dm=2a_1m-2a_1n\\\implies d\{n(m-1)-m(n-1)\}=2a_1(m-n)$

$\implies d(mn-n-mn+m)=2a_1(m-n)\\\implies d(m-n)=2a_1(m-n)\\\implies d = 2a_1\:--\:(1)$

Then, ratio of the mth to the nth term of the A.P

$=\frac{a_m}{a_n}\\=\frac{a_1+(m-1)d}{a_1+(n-1)d}\\=\frac{a_1+2a_1(m-1)}{a_1+2a_1(n-1)}\:\:\:[from\:(1)]\\=\frac{a_1(1+2m-2)}{a_1(1+2n-2)}\\=\frac{2m-1}{2n-1}$

Hence, shown