# 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.

S seema garhwal

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 $179:321$

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