Question

The ratio of the sum of n terms of two A.P.'s is (7n − 1): (3n + 11), Find the ratio of their 10^th terms.

Answer 1

Let a₁, a₂ be the first terms and d₁, d₂ the common differences of the two given AP.'s. Then, the sums of the its n terms are given by

Sn = \(\frac{n}{2}\)[2a₁ + (n − 1)d₁] and S_n = \(\frac{n}{2}\)[2a₂ + (n − 1)d₂]

∴ \(\frac{Sn}{Sn}\) = \(\frac{\frac{n}{2}[2a_1+(n-1)d_1]}{\frac{n}{2}[2a_2+(n-1)d_2]}\) = \(\frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}\)

It is given that

\(\frac{Sn}{Sn}\) = \(\frac{7n\, −\, 1}{3n\, + \,11}\)

\(\frac{Sn}{Sn}\) = \(\frac{7n\, −\, 1}{3n\, + \,11}\)

⇒ \(\frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_1}\) = \(\frac{7n\, −\, 1}{3n\, + \,11}\)            (ii)

To find the ratio of the 10^th ers of the two A.P.'s, we replace n by (2 × 10 − 1) i.e. "19’ in (i)

Replacing n ty "19’ in (i), we get

∴ \(\frac{2a_1+(19-1)d_1}{2a_2+(19-1)d_2}\) = \(\frac{7(19)\, −\, 1}{3(19)\, + \,11}\)

⇒ \(\frac{a_1+9d_1}{a_2+9d_2}\) = \(\frac{133\,−\,1}{ 57\, +\, 11} = \frac{132}{ 68}\)

⇒ \(\frac{a_1+9d_1}{a_2+9d_2}\) = ⇒ \(\frac{a_1+9d_1}{a_2+9d_2}\) = \(\frac{33}{17}\)

∴ Ratio of the 10^th terms of the two A.P.'s = \(\frac{33}{17}\)