Question

the ratio of the sums of m and n terms of an AP m2:n2 show that the ratio of the Mth and Nth terms are (2m-1):(2n-1)​

Answer 1

Required Answer:-

Question:

• The ratio of the sums of m and n terms of an AP is m² : n². Show that the ratio of the mth term and the nth term is (2m - 1) : (2n - 1)

Proof:

Given that,

Sum of m and n terms of an A.P. are in ratio m² : n²

Let the first term of the A.P. be a.

So,

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

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

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

$\sf \implies \dfrac{m }{n } = \dfrac{2a + md - d}{2a +nd- d }$

On cross multiplying, we get,

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

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

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

$\sf \implies 2am - 2an- md + nd = 0$

Now, factor out.

$\sf \implies 2a(m -n)- d(m - n) = 0$

$\sf \implies (2a - d)(m - n) = 0$

By zero product rule,

Either (2a - d) = 0 or (m - n) = 0

➡ 2a = d (but m ≠ n)

So,

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

As d = 2a,So,

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

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

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

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

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

(Hence Proved)

Important A.P formula:

1. Nth term of A.P = a + (n - 1)d
2. Sum of first n terms of A.P. = n/2[2a + (n - 1)d]

Answer 2

Answer:

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