Question

The sums of '$n$' terms of two arithmetic progression are in the ration $5n+4:9n+6$ , find the ration of their ${18}^{th}$ terms.

Answer 1

Let the two A.P s be
$a,(a+d),(a+2d)$.............(1)
Where, $a$ is first term and $d$ is common difference for first A.P
and $A(A+D).(A+2D)$, ....... (2)
Where, $A$ is first term and $D$ is common difference for second A.P
Given that ,
$\frac{\text{sum of the first n terms of}\left(1\right)}{\text{Sum of first n term of}\left(2\right)}=\frac{5n+4}{9n+6}$
$\Rightarrow \frac{\frac{n}{2}[2a+(n-1\left)d\right]}{\frac{n}{2}[2A+(n-1\left)D\right]}=\frac{5n+4}{9n+6}$
$\Rightarrow \frac{[2a+(n-1\left)d\right]}{[2A+(n-1\left)D\right]}=\frac{5n+4}{9n+6}$
$\Rightarrow \frac{[a+\frac{(n-1)}{2}d]}{[A+\frac{(n-1)}{2}D]}=\frac{5n+4}{9n+6}$......(i)
Since, the ratio of ${18}^{th}$ terms is $\frac{a_{18}}{A_{18}}=\frac{a+17d}{A+17D}$....(ii)
Now $\frac{[a+\frac{(n-1)}{2}d]}{[A+\frac{(n-1)}{2}D]}$ compare with $\frac{a+17d}{A+17D}$, we get
$\Rightarrow \frac{(n-1)}{2}=17$
$\Rightarrow n-1=34$
$\Rightarrow n=34+1$
$\therefore n=35$
Substitute $n=35$ in equation(i), we get
$\frac{[a+\frac{(35-1)}{2}d]}{[A+\frac{(35-1)}{2}D]}=\frac{5\left(35\right)+4}{9\left(35\right)+6}$
$\Rightarrow \frac{a+17d}{A+17D}=\frac{175+4}{315+6}$
$\therefore \frac{a_{18}}{A_{18}}=\frac{179}{321}$ [From equation(ii)]
Hence, the ratio of ${18}^{th}$ term of first A.P and ${18}^{th}$ term second A.P is $179:321$.