The sum of first n terms of two APs are in the ratio (3n + 8) : (7n + 15).

Question:

The sum of first n terms of two APs are in the ratio (3n + 8) : (7n + 15). Find the ratio of their 12th terms.

 

Solution:

Let the first term of the first AP be a
And common difference be d

$s_{n}=\frac{n}{2}[2 a+(n-1) d]$

And

$a_{n}=a+(n-1) d$

$a_{12}=a+11 d=3 n+8$

Similarly, for second A.P

Let first term be $A$

And common difference be $D$

$S_{n}=\frac{n}{2}[2 A+(n-1) D]$

And

$A_{n}=A+(n-1) D$

$A_{12}=A+11 D=7 n+15$

We have to find the ratio of 12 th term

$\frac{a_{12} \text { of first A.P }}{A_{12} \text { of second A.P }}=\frac{a+11 d}{A+11 D}$

$\frac{s_{n} \text { of first A.P }}{S_{n} \text { of second A.P }}=\frac{3 n+8}{7 n+15}$

$\frac{\frac{n}{2}[2 a+(n-1) d]}{\frac{n}{2}[2 A+(n-1) D]}=\frac{3 n+8}{7 n+15}$

$\frac{2 a+(n-1) d}{2 A+(n-1) D}=\frac{3 n+8}{7 n+15}$

$\Rightarrow \frac{a+\left(\frac{n-1}{2}\right) d}{A+\left(\frac{n-1}{2}\right) D}=\frac{a+11 d}{A+11 D}=\frac{3 n+8}{7 n+15}$       $\cdots(1)$

We have to find $\frac{a+11 d}{A+11 D}$

$\frac{n-1}{2}=11$

$\Rightarrow n=23$

Put $n=23$ in (1)

$\frac{a+\left(\frac{22}{2}\right) d}{A+\left(\frac{22}{2}\right) D}=\frac{69+8}{161+15}$

$\frac{a+11 d}{A+11 D}=\frac{77}{176}$

$\frac{a+11 d}{A+11 D}=\frac{7}{16}$

Ratio of 12 th term is $7: 16$

 

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