If the sums of n terms of two arithmetic progressions are in the ration $\Large\frac{3 n+5}{5 n+7}$, then their nth terms are in the ration.
A. $\begin{aligned}\frac{3 n-1}{5 n-1}\end{aligned}$ B. $\begin{aligned}\frac{3 n+1}{5 n+1}\end{aligned}$ C. $\begin{aligned}\frac{5 n+1}{3 n+1}\end{aligned}$ D. $\begin{aligned}\frac{5 n-1}{3 n-1}\end{aligned}$ Answer: Option BShow Answer
Solution(By Apex Team)
In first A.P. let its first term be a1 and common difference d1
and in second A.P., first term be a2 and common difference d2,
Then,
$\begin{aligned}\frac{S_n}{S_n}&=\frac{\frac{n}{2}\left[2a_1+(n-1)d_1\right]}{\frac{n}{2}\left[2a_2+(n-1)d_2\right]}\\
&=\frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}\\
&\therefore\frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}=\frac{3n+5}{5n+7}\end{aligned}$
Substituting n = 2n – 1, then
$\begin{aligned}&\frac{2a_1+(2n-2)d_1}{2a_2+(2n-2)d_2}=\frac{3(2n-1)+5}{5(2n-1)+7}\\
&\Rightarrow\frac{a_1+(n-1)d_1}{a_2+(n-1)d_2}=\frac{6n-3+5}{10n-5+7}\text{ (Dividing by 2) }\\
&\Rightarrow\frac{a_{1n}}{a_{2n}}=\frac{6n+2}{10n+2}\\
&\Rightarrow\frac{a_{1n}}{a_{2n}}=\frac{3n+1}{5n+1}\end{aligned}$
Related Questions On Progressions
How many terms are there in 20, 25, 30 . . . . . . 140?
A. 22B. 25
C. 23
D. 24
Find the first term of an AP whose 8th and 12th terms are respectively 39 and 59.
A. 5B. 6
C. 4
D. 3
Find the 15th term of the sequence 20, 15, 10 . . .
A. -45B. -55
C. -50
D. 0
The sum of the first 16 terms of an AP whose first term and third term are 5 and 15 respectively is
A. 600B. 765
C. 640
D. 680
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