Question

please answer the 23 question​

Answer 1

Answer:

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

Step-by-step explanation:

Let a be the first term and d be the common difference.

Given that ratio of sums of first m and n terms is m² : n².

∴ (Sum of first m terms)/(Sum of first n terms) = m²/n².

$=>\frac{\frac{m}{2}[2a+(m-1)*d]}{\frac{n}{2}[2a+(n- 1)*d]}=\frac{m^2}{n^2}$

$=>\frac{2a + (m - 1) * d}{2a + (n - 1) * d} = \frac{m}{n}$

$=>[2a+(m - 1)*d]n=[2a+(n-1)*d]m$

$=>2an + (m - 1)dn=2am+(n - 1)dm$

$=>d[(m - 1) * n - (n - 1) * m] = 2am - 2an$

$=> d(mn - n - nm + m) = 2a(m - n)$

$=> d(m - n) = 2a(m - n)$

$=> d = 2a$

Now,

Ratio of mth term and nth term is:

$=>\frac{a_{m} }{a_{n}}$

$=>\frac{a+(m-1) * 2a}{a + (n - 1) * 2a}$

$=>\frac{a(1 + 2m - 2)}{a(1 + 2n - 2)}$

$=>\frac{1+2m-2}{1+2n-2}$

$=>\frac{2m-1}{2n-1}$

Therefore,ratio of its mth term and nth term is 2m - 1 : 2n - 1

Hope it helps!