The sums of n terms of two arithmetic progres | vedclass

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(A) Let $a_1, a_2$ and $d_1, d_2$ be the first terms and common differences of the first and second arithmetic progressions,respectively.The sum of $n

Vedclass · Accepted Answer

$179 : 321$

Vedclass · Answer

(A) Let $a_1, a_2$ and $d_1, d_2$ be the first terms and common differences of the first and second arithmetic progressions,respectively.The sum of $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.According to the given condition:$\frac{S_{n,1}}{S_{n,2}} = \frac{\frac{n}{2}[2a_1 + (n-1)d_1]}{\frac{n}{2}[2a_2 + (n-1)d_2]} = \frac{5n+4}{9n+6}$$\Rightarrow \frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{5n+4}{9n+6}$We need to find the ratio of the $18^{th}$ terms,which is $\frac{a_1 + 17d_1}{a_2 + 17d_2}$.To get the coefficient $17$ for $d_1$ and $d_2$,we set $\frac{n-1}{2} = 17$,which gives $n-1 = 34$,so $n = 35$.Substituting $n = 35$ in the ratio:$\frac{2a_1 + 34d_1}{2a_2 + 34d_2} = \frac{5(35)+4}{9(35)+6}$$\Rightarrow \frac{2(a_1 + 17d_1)}{2(a_2 + 17d_2)} = \frac{175+4}{315+6}$$\Rightarrow \frac{a_1 + 17d_1}{a_2 + 17d_2} = \frac{179}{321}$Thus,the ratio of their $18^{th}$ terms is $179 : 321$.

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

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