The sum of n terms of two arithmetic progressions are in the ratio (3n + 8) : (7n + 15). Find the ratio of their 12th terms
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The sum of n terms of two arithmetic progressions are in the ratio (3n + 8) : (7n + 15). Find the ratio of their 12th terms
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[tex]\sf\red{\underline{\underline{Answer:}}}[/tex]
[tex]\sf{The \ ratio \ of \ 12^{th} \ term \ of \ both \ the}[/tex]
[tex]\sf{arithmetic \ progressions \ is \ 7:16.}[/tex]
[tex]\sf\orange{Given:}[/tex]
[tex]\sf{\implies{The \ sum \ of \ n \ terms \ of \ the \ two}}[/tex]
[tex]\sf{arithmetic \ progressions \ are \ in \ the \ ratio}[/tex]
[tex]\sf{(3n+8) \ : \ (7n+15).}[/tex]
[tex]\sf\pink{To \ find:}[/tex]
[tex]\sf{The \ ratio \ of \ their \ 12^{th} \ terms.}[/tex]
[tex]\sf\green{\underline{\underline{Solution:}}}[/tex]
[tex]\boxed{\sf{t_{n}=a+(n-1)d}}[/tex]
[tex]\sf{For \ first \ AP,}[/tex]
[tex]\sf{Let, \ First \ term=a_{1} \ and \ Common \ difference=d_{1}}[/tex]
[tex]\sf{\therefore{12^{th} \ term \ is \ 1t_{12}}}[/tex]
[tex]\sf{\underline{\underline{\therefore{1t_{12}=a_{1}+11d_{1}...(1)}}}}[/tex]
[tex]\sf{For \ second \ AP,}[/tex]
[tex]\sf{Let, \ First \ term=a_{2} \ and \ Common \ difference=d_{2}}[/tex]
[tex]\sf{\therefore{12^{th} \ term \ is \ 2t_{12}}}[/tex]
[tex]\sf{\underline{\underline{\therefore{2t_{12}=a_{2}+11d_{2}...(2)}}}}[/tex]
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[tex]\sf{Now, \ Let \ the \ sum \ of \ n \ terms \ of two \ terms}[/tex]
[tex]\sf{of \ two \ arithmetic \ progressions \ be \ 1S_{n}}[/tex]
[tex]\sf{and \ 2S_{2} \ respectively.}[/tex]
[tex]\boxed{\sf{S_{n}=\frac{n}{2}[2a+(n-1)d]}}[/tex]
[tex]\sf{According \ to \ the \ given \ condition.}[/tex]
[tex]\sf{\frac{1S_{1}}{2S_{2}}=\frac{3n+8}{7n+15}}[/tex]
[tex]\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}}}[/tex]
[tex]\sf{\therefore{\frac{2a_{1}+(n-1)d_{1}}{2a_{2}+(n-1)d_{2}}=\frac{3n+8}{7n+15}}}[/tex]
[tex]\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}}}[/tex]
[tex]\sf{\therefore{\frac{a_{1}+\frac{(n-1)d_{1}}{2}}{a_{2}+\frac{(n-1)d_{2}}{2}}=\frac{3n+8}{7n+15}}}[/tex]
[tex]\sf{Put \ n=23, \ we \ get}[/tex]
[tex]\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}}[/tex]
[tex]\sf{\therefore{\frac{a_{1}+\frac{22d_{1}}{2}}{a_{2}+\frac{22d_{2}}{2}}=\frac{69+8}{161+15}}}[/tex]
[tex]\sf{\therefore{\frac{a_{1}+11d_{1}}{a_{2}+11d_{2}}=\frac{77}{176}}}[/tex]
[tex]\sf{\underline{\underline{\therefore{\frac{a_{1}+11d_{1}}{a_{2}+11d_{2}}=\frac{7}{16}...(3)}}}}[/tex]
[tex]\sf{...from \ (1), \ (2) \ and \ (3)}[/tex]
[tex]\sf{\frac{1t_{12}}{2t_{12}}=\frac{7}{16}}[/tex]
[tex]\sf\purple{\tt{\therefore{The \ ratio \ of \ 12^{th} \ term \ of \ both \ the}}}[/tex]
[tex]\sf\purple{\tt{arithmetic \ progressions \ is \ 7:16.}}[/tex]
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