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

$\huge\underline\mathbb{\pink Q\pink {U} \pink {E} \pink {ST} \pink {I}\pink {ON???}}$

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).

$\huge\underline\mathbb{\red S\pink {0} \purple {L} \blue {UT} \orange {1}\green {ON :}}$

• Let a be the first term and
• d be the common difference of given AP.

$\sf\underline{\underline{\green{Then, \: }}}$

Sₘ = sum of first m terms

Sₙ = sum of first n terms

$\sf\underline{\underline{\green{<strong>Now,</strong> \: }}}$

$\sf \frac{S_m}{S_n} = \frac{ {m}^{2} }{ {n}^{2} } \\ \\ \implies \sf \frac{ \dfrac{m}{2}[2a - (m - 1)d] }{\dfrac{n}{2}[2a - (n - 1)d] } = \frac{m {}^{2} }{n {}^{2} } \\ \\ \implies \sf \frac{2a + (m - 1)d}{2a + (n - 1)d} = \frac{m}{n} \\ \\ \bf <strong>on</strong> \: <strong>cross - multiplying</strong> : \\ \\ \sf n \{2a + (m - 1)d \} = m \{2a + (n - 1)d \} \\ \\ \implies \sf 2an + mnd - nd = 2am + mnd - md \\ \\ \implies \sf 2an - 2am = nd - md \\ \\ \implies \sf 2a(n - m) = d(n - m) \\ \\ \implies \sf 2a = d$

$\sf\underline{\underline{\green{<strong>Therefore</strong>, \: }}}$

Tₘ = mᵗʰ term of the AP

Tₙ = nᵗʰ term of the AP

$\sf \therefore \: \frac{T_m}{T_n} = \frac{a + (n - 1)d}{a + (m - 1)d} \\ \\ \rightarrow \sf \frac{a + (m - 1) \: . \: 2a}{a + (n - 1) \:. \: 2a} \\ \\ \rightarrow \sf \frac{a + 2am - 2a}{a + 2an - 2a} \\ \\ \rightarrow \sf \frac{2am - a}{2an - a} \\ \\ \rightarrow \sf \frac{a(2m - 1)}{a(2n - 1)} \\ \\ \<strong>rightarrow</strong> \sf \frac{2m - 1}{2n - 1} \\ \\\boxed{\bf \therefore \: <strong>T_m: T_n = 2m - 1: 2n - 1</strong>}$

Answer 2

Given :

• Ratio of the sum of first m and n terms of an A.P. is m² : n²

To prove :

• The ratio of its $\sf{m_{th}}$ and $\sf{n_{th}}$ terms is (2 m - 1) : (2 n - 1)

proof :

• Let, first term of AP be a and common difference be d

As given that,

• Ratio of the sum of first m and n terms of an A.P. is m² : n²

so,

$\implies\sf{\dfrac{S_m}{S_n}=\dfrac{m^2}{n^2}}$

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

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

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

cross multiplying

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

$\implies\sf{2an+nd(m-1)=2am+dm(n-1)}$

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

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

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

$\implies\sf{\red{2a=d}\;\;\;\;....equation (1)}$

Now,

Ratio of $\sf{m_{th}\;and\:n_{th}}$ terms will be,

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

using equation (1)

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

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

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

$\implies\red{\sf{\dfrac{a_m}{a_n}=\dfrac{2m-1}{2n-1}}}$

Proved .