Question

If the ratio of sum of n term of two A.P's is (3n + 8) : (7n + 15), then the ratio of ${12}^{th}$ terms is-

• A. 16: 7
• B. 7: 16
• C. 7: 12
• D. 12: 5

Answer 1

The correct option is B 7: 16

For $1^{st}$ AP

Let first term be a common difference be d

sum of n terms $s_n=\frac{n}{2}(2a+(n-1\left)d\right)$

$n^{th}$ term $a_n=a(n-1)d$

Similarly for $2^{nd}$ AP

Let first term $=$A common difference be D

$S_n=\frac{n}{2}(2A+(n-1\left)D\right)$ & $n^{th}$ term $=A_n=A+(n-1)D$

We need ratio of ${12}^{th}$ term

i.e., $\frac{a_{12}offirstAP}{A_{12}ofsecondAP}$

$=\frac{a+(12-1)d}{A+(12-1)D}$

$=\frac{a+11d}{a+11D}$

It is given that

$\frac{Sumofntermsof{1}^{st}AP}{Sumofntermsof{2}^{nd}AP}=\frac{3n+8}{7n+15}$

$\therefore \frac{2a+(n-1)d}{2A+(n-1)D}=\frac{3n+8}{7n+15}$ ………….$\left(1\right)$

$\frac{2(a+\left(\frac{n-1}{2}\right)d)}{2(A+\left(\frac{n-1}{2}\right)D)}=\frac{3n+8}{7n+15}$ ………………$\left(1\right)$

we need to find $\frac{a+11d}{A+11D}$

Hence $\frac{n-1}{2}=11$

$n-1=22$

$n=23$

Putting $n=23$ in $\left(1\right)$

$\frac{a+\left(\frac{23-1}{2}\right)d}{A+\left(\frac{23-1}{2}\right)D}=\frac{3\times 23+18}{7\times 23+15}$

$\therefore \frac{a+11d}{A+11D}=\frac{7}{16}$

Hence ratio of their ${12}^{th}$ term is $\frac{7}{16}$ i.e., $7:16$.