Question

The ratio of sum of 'm' terms and 'n' terms of an arithmetic

sequence is m2 : n2

.Then the ratio of the mth and nth term is

A) m : n B) (2m+1) : (2n+1)

C) m2 : n2 D) (2m –1) : (2n –1)​

Answer 1

Answer:

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

Step-by-step explanation:

Let Sₘ and Sₙ be the sum of 'm' terms and 'n' terms of AP respectively.

Then,

Sₘ/Sₙ = m²/n²

♦ S = Total number of terms/2 [2a + (m - 1)d]

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

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

$\\ \\ \implies \sf n [2a + (m - 1)d] = m [2a + (n - 1)d]$

$\\ \\ \implies \sf 2an + mnd - nd = 2am + nmd - md$

$\\ \\ \implies \sf md - nd = 2am - 2an$

$\\ \\ \implies \sf (m - n)d = 2a(m - n)$

$\\ \\ \implies \sf d = 2a$

Finally, find the ratio of the mth and nth term:

$\sf \implies \dfrac{t_m}{t_n} = \dfrac{a + (m - 1)d}{a + (n - 1)d}$

$\\ \\ \sf = \dfrac{a + (m - 1)2a}{a + (n - 1)2a}$

$\\ \\ \sf = \dfrac{a + 2am - 2a}{a + 2an - 2a}$

$\\ \\ \sf = \dfrac{a(1 + 2m - 2)}{a(1 + 2n - 2)}$

$\\ \\ \sf = \dfrac{1 + 2m - 2}{1 + 2n - 2}$

$\\ \\ \sf = \dfrac{2m - 1}{ 2n - 1}$

$\\ \\ \purple {\sf = 2m - 1 \ratio 2n - 1}$

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Answer 2

Answer:

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

Step-by-step explanation:

hope you understand that