The sums of n terms of two arithmetic progressions are in the ratio 5n + 4: 9n + 6. Find the ratio of their 18th terms.
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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6 Comment(s) on this Question
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)?
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