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

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

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Given: The sums of n terms of two arithmetic progressions are in the ratio.5n + 4 : 9n + 6

There are two AP's with first terms =a_1,a_2    and common difference =  d_1,d_2

\Rightarrow \, \, \frac{\frac{n}{2}[2a_1+(n-1)d_1]}{\frac{n}{2}[2a_2+(n-1)d_2]}=\frac{5n+4}{9n+6}

\Rightarrow \, \, \frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}=\frac{5n+4}{9n+6}

Substituting n=35,we get

\Rightarrow \, \, \frac{2a_1+(35-1)d_1}{2a_2+(35-1)d_2}=\frac{5(35)+4}{9(35)+6}

\Rightarrow \, \, \frac{2a_1+34 d_1}{2a_2+34d_2}=\frac{5(35)+4}{9(35)+6}

\Rightarrow \, \, \frac{a_1+17 d_1}{a_2+17d_2}=\frac{179}{321}

\Rightarrow \, \, \frac{18^t^h \, term \, of\, first \, AP}{18^t^h\, term\, of\, second\, AP}=\frac{179}{321}

Thus, the ratio of the 18th term of AP's is 179:321

Posted by

seema garhwal

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