Question

The ratio of sums of m and n terms of an AP is m^2 : n^2. Show that the ratio of the m'th and n'th terms is (2m-1) : (2n-1)

Answer 1

≡QUESTION≡

The ratio of sums of m and n terms of an AP is m² : n². Show that the ratio of the m'th and n'th terms is (2m-1) : (2n-1).

                                                     

║⊕ANSWER⊕║

Sum of m terms of an A.P. = m/2 [2a + (m -1)d] = m²

Sum of n terms of an A.P. = n/2 [2a + (n -1)d] = n²

$m/2 [2a + (m -1)d] / n/2 [2a + (n -1)d] = m^2:n^2\\ \\2an + mnd - nd = 2am + mnd - md\\ 2an - 2am = nd - md\\2a (n -m) = d(n - m)\\ 2a = d$  

(∵ m- n get cancelled)

Ratio of m th term to nth term:

$[a + (m - 1)d] / [a + (n - 1)d]\\ [a + (m - 1)2a] / [a + (n - 1)2a]\\ a [1 + 2m - 2] / a[1 + 2n -2]\\(2m - 1) / (2n -1)$

∴ The ratio of the m'th and n'th terms is (2m-1) : (2n-1)