Question

The ratio of the sum of first m and n terms of an Ap is m^2:n^2.show that the ratio of its nth and nth term is (2m-1):(2n-1)​
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Answer 1

GIVEN :–

• The ratio of the sum of first m and n terms of an Ap is m² : n².

TO PROVE :–

• The ratio of its mth and nth term is (2m-1):(2n-1).

SOLUTION :–

• We know that Sum of an A.P. is –

$\\ \dashrightarrow \large{ \boxed { \bold{S_{n} = \dfrac{n}{2} \{2a + (n - 1)d \}}}}$

• Here –

$\\ \: \: {\huge{.}} \: \: { \bold{a = first \: \: term}}$

$\\ \: \: {\huge{.}} \: \: { \bold{d = common \: \: difference}}$

• According to the question –

$\\ \implies { \bold{ \dfrac{S_{m}}{S_{n}} = \dfrac{m^2}{n^2}}}$

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

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

$\\ \implies { \bold{ {n(2a + md - d)} = {m(2a +nd - d)}}}$

$\\ \implies { \bold{2an + mnd - nd = 2am + mnd - md}}$

$\\ \implies { \bold{2an - nd = 2am - md}}$

$\\ \implies { \bold{2an - 2am = nd - md}}$

$\\ \implies { \bold{2a(n -m) = d(n - m)}}$

$\\ \implies { \bold{2a= d \: \: \: \: - - - - eq.(1)}}$

• We know that nth term –

$\\ \dashrightarrow \large{ \boxed { \bold{T_{n} = a + (n - 1)d }}}$

$\\ \implies { \bold{ \dfrac{T_{m}}{T_{n}} = \dfrac{a + (m- 1)d}{a + (n - 1)d}}}$

• Using eq.(1) –

$\\ \implies { \bold{ \dfrac{T_{m}}{T_{n}} = \dfrac{a + (m- 1)2a}{a + (n - 1)2a}}}$

$\\ \implies { \bold{ \dfrac{T_{m}}{T_{n}} = \dfrac{a \{1+ (m- 1)2 \}}{a \{1+ (n - 1)2 \}}}}$

$\\ \implies { \bold{ \dfrac{T_{m}}{T_{n}} = \dfrac{1+ (m- 1)2} {1+ (n - 1)2 }}}$

$\\ \implies { \bold{ \dfrac{T_{m}}{T_{n}} = \dfrac{1+2m - 2} {1+2n - 2 }}}$

$\\ \implies \large {\boxed{ { \bold{ \dfrac{T_{m}}{T_{n}} = \dfrac{2m -1} {2n -1}}}}}$

$\\ \: \: \: \: \: \: \: \: \: \: \: \large {\boxed{{ \underline{ \bold{Hence \: \: proved }}}}}$