Question

19. The ratio of the sums of first m and first 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

Step-by-step explanation:

refer to the photos attached

Answer 2

Solution:

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

Sₘ = sum of first m terms

Sₙ = sum of first n terms

Now,

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

Therefore,

Tₘ = mᵗʰ term of the AP

Tₙ = nᵗʰ term of the AP

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