The sum of n terms of two A.P.’s are in the ratio 5n + 4 : 9n + 6. Then, the ratio of their 18th term is
A. $\frac{179}{321}$ B. $\frac{178}{321}$ C. $\frac{175}{321}$ D. $\frac{176}{321}$ Answer: Option AShow Answer
Solution(By Apex Team)
Let a1, d2 be the first terms of two ratios S and S’ and d1, d2 be their common difference respectively
Then,
$\begin{aligned}&S_n=\frac{n}{2}\left[2a_1+(n-1)d_1\right]\text{ and }\\
&S_n^{\prime}=\frac{n}{2}\left[2a_2+(n-1)d_2\right]\\
&\text{ Now }\\
&\frac{S_n}{S_n^{\prime}}=\frac{\frac{n}{2}\left[2a_1+(n-1)d_1\right]}{\frac{n}{2}\left[2a_2+(n-1)d_2\right]}\\
&=\frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}\\
&\text{ But }\frac{S_n}{S_n^{\prime}}=\frac{5n+4}{9n+6}\\
&\therefore\frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}=\frac{5n+4}{9n+6}\end{aligned}$
Now we have to find the ratios in 18th term
Here n = 18
$\begin{aligned}\therefore\frac{2a_1+(18-1)d_1}{2a_2+(18-1)d_2}&=\frac{5(2n-1)+4}{9(2n-1)+6}\\
=&\frac{5(2\times18-1)+4}{9(2\times18-1)+6}\\
=&\frac{5\times35+4}{9\times35+6}\\
=&\frac{175+4}{315+6}\\
=&\frac{179}{321}\end{aligned}$
Related Questions On Progressions
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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 . . .
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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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