Question

The sums of n terms of two arithmetic progressions are in the ratio 5n+4 : 9n+6. Find the ratio of their 18th terms.

Answer 1

There are two AP with different first term and common
difference.
For the first AP
Let first term be $a$
Common difference $=d$
Sum of $n$ terms $=$
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
and $n^{th}$ term $=a_n=a+(n-1)d$
For the second AP
Let first term be $A$
Common difference $=D$
Sum of $n$ terms
$=S_n=\frac{n}{2}[2A+(n-1\left)D\right]$
and $n^{th}$ term $=A_n=A+(n-1)D$
It is given that
$\frac{SumofntermsoffirstA.P}{SumofntermsofsecondA.P}=\frac{5n+4}{9n+6}$
$\Rightarrow \frac{\frac{n}{2}[2a+(n-1\left)d\right]}{\frac{n}{2}[2A+(n-1\left)D\right]}=\frac{5n+4}{9n+6}$
$\Rightarrow \frac{n[a+\left(\frac{n-1}{2}\right)d]}{n[A+\left(\frac{n-1}{2}\right)D}=\frac{5n+4}{9n+6}$
$\Rightarrow \frac{a+\left(\frac{n-1}{2}\right)d}{A+\left(\frac{n-1}{2}\right)D}=\frac{5n+4}{9n+6}...\left(i\right)$
Now, we need to find ratio of their ${18}^{th}$ term
i.e. $\frac{{18}^{th}termoffirstAP}{{18}^{th}termofsecondAP}$
$=\frac{a_{18}offirstAP}{A_{18}ofsecondAP}$
$=\frac{a+(18-1)d}{A+(18-1)D}$
$=\frac{a+17d}{A+17D}$
Hence, $\frac{n-1}{2}=17$
$n-1=17\times 2$
$\Rightarrow n=34+1$
$\Rightarrow n=35$
Putting $n=35$ in equation.$\left(i\right),$ we get
$\frac{a+\left(\frac{35-1}{2}\right)d}{A+\left(\frac{35-1}{2}\right)D}=\frac{5\left(35\right)+4}{9\left(35\right)+6}$
$\Rightarrow \frac{a+17d}{A+17D}=\frac{175+4}{315+6}$
$\Rightarrow \frac{{18}^{th}termoffirstAP}{{18}^{th}termofsecondAP}=\frac{179}{321}$
Hence the ratio of ${18}^{th}$ term of $1^{st}$ AP and ${18}^{th}$ term if $2^{nd}$ AP is
$179:321$