Let a1, a2, and d1, d2 be the first terms and the common difference of the first and second arithmetic progression respectively.
According to the given condition,
\begin{align} \frac{Sum \;of \;n \;terms \;of \;first\; A.P.}{Sum\; of \;n\; terms \;of \;second \;A.P.} = \frac{5n+4}{9n+6} \end{align}
\begin{align} ⇒\frac{\frac{n}{2}\left[2a_1 + (n-1)d_1\right]}{\frac{n}{2}\left[2a_2 + (n-1)d_2\right]} = \frac{5n+4}{9n+6} \end{align}
\begin{align} ⇒\frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{5n+4}{9n+6} \;\;\;\;...(1)\end{align}
Substituting n = 35 in (1), we obtain
\begin{align} ⇒\frac{2a_1 + 34d_1}{2a_2 + 34d_2} = \frac{5(35)+4}{9(35)+6} \end{align}
\begin{align} ⇒\frac{a_1 + 17d_1}{a_2 + 17d_2} = \frac{179}{321} \;\;\;\;...(2)\end{align}
\begin{align} \frac{18^{th} \;term \;of\; first\; A.P.}{18^{th} \;term \;of\; second\; A.P.}=\frac{a_1 + 17d_1}{a_2 + 17d_2} \;\;\;\;...(3)\end{align}
From (2) and (3), we obtain
\begin{align} \frac{18^{th} \;term \;of\; first\; A.P.}{18^{th} \;term \;of\; second\; A.P.}=\frac{179}{321}\end{align}
Thus, the ratio of 18th term of both the A.P.s is 179: 321.
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Evaluate
We take 35 because. . If the ratio of the sum of n terms of ap is given, then to find the ratio of their nth terms, we replace n by (2n-1) in the ratio of the sums of n terms. 2*18-1 = 35
TO OBTAIN THE 18TH TERMS (n-1)/2=18-1 (n-1)/2=17 n-1=34 n=34+1 n=35
here in question we need to find 18th term, now 18+18-1=35. that's why n=35
Why do we take n=35 in (1)?pls tell me
(n-1)/2=An-1
Why do we take n=35 in (1)?