Question

The sum of n terms of two arithmetic progressions are in the ratio (3n + 8) : (7n + 15). Find the ratio of their 12th terms

Answer 1

$\sf\red{\underline{\underline{Answer:}}}$

$\sf{The \ ratio \ of \ 12^{th} \ term \ of \ both \ the}$

$\sf{arithmetic \ progressions \ is \ 7:16.}$

$\sf\orange{Given:}$

$\sf{\implies{The \ sum \ of \ n \ terms \ of \ the \ two}}$

$\sf{arithmetic \ progressions \ are \ in \ the \ ratio}$

$\sf{(3n+8) \ : \ (7n+15).}$

$\sf\pink{To \ find:}$

$\sf{The \ ratio \ of \ their \ 12^{th} \ terms.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\boxed{\sf{t_{n}=a+(n-1)d}}$

$\sf{For \ first \ AP,}$

$\sf{Let, \ First \ term=a_{1} \ and \ Common \ difference=d_{1}}$

$\sf{\therefore{12^{th} \ term \ is \ 1t_{12}}}$

$\sf{\underline{\underline{\therefore{1t_{12}=a_{1}+11d_{1}...(1)}}}}$

$\sf{For \ second \ AP,}$

$\sf{Let, \ First \ term=a_{2} \ and \ Common \ difference=d_{2}}$

$\sf{\therefore{12^{th} \ term \ is \ 2t_{12}}}$

$\sf{\underline{\underline{\therefore{2t_{12}=a_{2}+11d_{2}...(2)}}}}$

________________________________

$\sf{Now, \ Let \ the \ sum \ of \ n \ terms \ of two \ terms}$

$\sf{of \ two \ arithmetic \ progressions \ be \ 1S_{n}}$

$\sf{and \ 2S_{2} \ respectively.}$

$\boxed{\sf{S_{n}=\frac{n}{2}[2a+(n-1)d]}}$

$\sf{According \ to \ the \ given \ condition.}$

$\sf{\frac{1S_{1}}{2S_{2}}=\frac{3n+8}{7n+15}}$

$\sf{\therefore{\frac{\frac{n}{2}[2a_{1}+(n-1)d_{1}]}{\frac{n}{2}[2a_{2}+(n-1)d_{2}]}=\frac{3n+8}{7n+15}}}$

$\sf{\therefore{\frac{2a_{1}+(n-1)d_{1}}{2a_{2}+(n-1)d_{2}}=\frac{3n+8}{7n+15}}}$

$\sf{\therefore{\frac{2[a_{1}+\frac{(n-1)d_{1}}{2}]}{2[a_{2}+\frac{(n-1)d_{2}}{2}]}=\frac{3n+8}{7n+15}}}$

$\sf{\therefore{\frac{a_{1}+\frac{(n-1)d_{1}}{2}}{a_{2}+\frac{(n-1)d_{2}}{2}}=\frac{3n+8}{7n+15}}}$

$\sf{Put \ n=23, \ we \ get}$

$\sf{\frac{a_{1}+\frac{(23-1)d_{1}}{2}}{a_{2}+\frac{(23-1)d_{2}}{2}}=\frac{3(23)+8}{7(23)+8}}$

$\sf{\therefore{\frac{a_{1}+\frac{22d_{1}}{2}}{a_{2}+\frac{22d_{2}}{2}}=\frac{69+8}{161+15}}}$

$\sf{\therefore{\frac{a_{1}+11d_{1}}{a_{2}+11d_{2}}=\frac{77}{176}}}$

$\sf{\underline{\underline{\therefore{\frac{a_{1}+11d_{1}}{a_{2}+11d_{2}}=\frac{7}{16}...(3)}}}}$

$\sf{...from \ (1), \ (2) \ and \ (3)}$

$\sf{\frac{1t_{12}}{2t_{12}}=\frac{7}{16}}$

$\sf\purple{\tt{\therefore{The \ ratio \ of \ 12^{th} \ term \ of \ both \ the}}}$

$\sf\purple{\tt{arithmetic \ progressions \ is \ 7:16.}}$

Answer 2

Refers to attachment ♡~