Question

The ratio of the sums of m and n terms of an AP is m²:n². Find the ratio of m,th and n,th term is (2m-1):(2n-1)​

Answer 1

EXPLANATION.

Ratio of the sum of the m and n terms of an

Ap = m²:n².

To find ratio of mth and nth

(2m - 1) : ( 2n - 1).

Sum of Nth terms of an Ap.

$\sf : \implies \: s_{n} \: = \dfrac{n}{2}(2a \: + \: (n - 1)d)$

$\sf : \implies \: s_{m} \: = \dfrac{m}{2} (2a \: + (m - 1)d) = {m}^{2} \\ \\ \sf : \implies \: s_{n} \: = \dfrac{n}{2} (2a \: + (n - 1)d) = {n}^{2}$

$\sf : \implies \: \dfrac{ \dfrac{m}{2} (2a + (m - 1)d)}{ \dfrac{n}{2} (2a + (n - 1)d)} = \dfrac{ {m}^{2} }{ {n}^{2} } \\ \\ \sf : \implies \: \: \frac{2a + (m - 1)d}{2a + (n - 1)d} = \frac{m}{n} \\ \\ \sf : \implies \: n[2a + (m - 1)d] = m[2a + (n - 1)d]$

$\sf : \implies \: n[2a + md - d] = m[2a + nd - d \\ \\ \sf : \implies \: 2an \: + mnd \: - nd \: = 2am \: + mnd \: - md \\ \\ \sf : \implies \: md \: - \: nd \: = 2am \: - \: 2an \\ \\ \sf : \implies \: (m - n)d \: = 2a(m - n) \\ \\ \sf : \implies \: d \: = 2a$

$\sf : \implies \: \dfrac{ s_{m} }{ s_{n} } = \dfrac{2a \: + \: (m \: - 1)d}{2a \: + \: (n - 1)d} \\ \\ \sf : \implies \: \dfrac{ s_{m} }{ s_{n} } \: = \frac{2a \: + \: (m - 1)2a}{2a \: +(n - 1)2a } \\ \\ \sf : \implies \: \dfrac{ s_{m} }{ s_{n} } = \frac{a(1 + 2m - 2)}{a(1 + 2n - 2)} \\ \\ \sf : \implies \: \dfrac{ s_{m} }{ s_{n} } = \frac{2m - 1}{2n - 1}$

Answer 2

✯Given✯

• Ratio of the sums of m and n terms of an AP is m²:n².

✯ To find ✯

• The ratio of m,th and n,th term is (2m-1):(2n-1)

꧁ Solution ꧂

We know that :-

$\\ {\sf{\implies{S_n = \frac{n}{2}(2a + (n - 1)d) }}}$

Where $S_n$= sum of nth term of A.P.

• n = number if terms.
• a = first term
• d = common difference

Thus, Sum of n term = $S_n$

$\\ {\sf{\implies{S_n = \frac{n}{2}(2a + (n - 1)d) }}}$

and also

Sum of n term = $S_m$

$\\ {\sf{\implies{S_m = \frac{m}{2}(2a + (m - 1)d) }}}$

It is given that :-

Ratio of sums of m and n terms of an A.P is m² : n².

${\large{\bf{\implies{ \frac{ Sum \: of \: m \: terms} {Sum \: of \: n \: terms }= \frac{ m²}{n²}}}}}$

${\large{\sf{\implies{ \frac{ S_m}{S_n }= \frac{m²}{ n²}}}}}$

${\large{\sf{\implies{ \frac{ \frac{m}{2} (2a + (m - 1)d)}{ \frac{n}{2}(2a + (n - 1)d) } = \frac{ {m}^{2} }{ {n}^{2} } }}}}$

${\large{\sf{\implies{ \frac{ m (2a + (m - 1)d)}{ n(2a + (n - 1)d) } = \frac{ {m}^{2} }{ {n}^{2} } }}}}$

${\large{\sf{\implies{ \frac{ 2a + (m - 1)d}{ 2a + (n - 1)d } = \frac{n}{m} \times \frac{ {m}^{2} }{ {n}^{2} } }}}}$

${\large{\bf{\implies{ \frac{ 2a + (m - 1)d}{ 2a + (n - 1)d } = \frac{m}{n} }}}} \: \: \: \: \: \: \: \: ......(1)$

We need to show that the ratio of m and n term is (2m - 1) : (2n - 1) :-

Finding n^th and m^th terms :-

We know that

${\bf{\implies{a_n = a + ( n -1 )d }}}$

${ \sf{So, n^{nt} \: term = a_n }} \\ \: \: \: \: \: \: \: \: \: \: \: \: { \sf{ = a + (n - 1)d}}$

Similarly :-

${ \sf{So, m^{nt} \: term = a_m }} \\ \: \: \: \: \: \: \: \: \: \: \: \: { \sf{ = a + (m- 1)d}}$

We need to show that the ratio of m and n term is (2m - 1) : (2n - 1) :-

${\bf{\implies{ \frac{ {m}^{th} \: term}{ {n}^{th} \: term} = \frac{(2m - 1)}{(2n - 1)} }}}$

${\bf{\implies{ \frac{a + (m - 1)d}{ a + (n - 1)d} = \frac{(2m - 1)}{(2n - 1)} }}} \: \: .......(2)$

From 1 :-

${\large{\bf{\implies{ \frac{ 2a + (m - 1)d}{ 2a + (n - 1)d } = \frac{m}{n} }}}} \: \: \: \: \: \: \: \:$

Replacing m with 2m -1 and n with 2n -1

${\bf{\implies{ \frac{ 2a + ((2m - 1) - 1)d}{ 2a + ((2n - 1) - 1)d} = \frac{(2m - 1)}{(2n - 1)} }}}$

${\bf{\implies{ \frac{ 2a + (2m - 1- 1)d}{ 2a + (2n - 1- 1)d} = \frac{(2m - 1)}{(2n - 1)} }}}$

${\bf{\implies{ \frac{ 2a + (2m - 2)d}{ 2a + (2n - 2)d} = \frac{(2m - 1)}{(2n - 1)} }}}$

${\bf{\implies{ \frac{ 2a + 2(m - 1)d}{ 2a +2 (n - 1)d} = \frac{(2m - 1)}{(2n - 1)} }}}$

${\bf{\implies{ \frac{ 2(a + (m - 1)d)}{ 2(a +2 (n - 1)d)} = \frac{(2m - 1)}{(2n - 1)} }}}$

${\bf{\implies{ \frac{ a + (m - 1)d}{ a +2 (n - 1)d} = \frac{(2m - 1)}{(2n - 1)} }}}$

${ \underline{ \boxed{ \blue{\bf{\implies{ \frac{ {m}^{th} \: term \: of \: A.P }{ {n}^{th} \: term \: of A.P } = \frac{(2m - 1)}{(2n - 1)} }}}}}}$

Thus, ratio of mth and nth term of A.P is 2m -1 : 2m - 1

Hence, proved.