Question

The sum of n terms of two ap are in ratio (5n+4) : (9n+6) Find the ratio of their 18th term?

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Answer 1

◆ Arithmetic Progressions ◆

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Answer 2

Answer:

The ratio of $18^{\text {th }}$ term of both the AP is 179:321

Solution:

Let us assume that for the first AP, the first term is a and common difference is d

For the second AP, the first term is A and the common difference is D,

Now as per the problem,

$\frac{\frac{n}{2}[2 a+(n-1) d]}{\frac{n}{2}[2 A+(n-1) D]}=\frac{5 n+4}{9 n+6}$

$\frac{[2 a+(n-1) d]}{[2 A+(n-1) D]}=\frac{5 n+4}{9 n+6}$

$\frac{\left[a+\frac{(n-1) d}{2}\right]}{\left[A+\frac{(n-1) D}{2}\right]}=\frac{5 n+4}{9 n+6}$……………….. (i)

Now the ratio of 18th term of both the ap is $=\frac{a+17 d}{A+17 D}$….. (ii)

Hence $\frac{n-1}{2}=17$

n=35

Now putting value of n in equation (i), we get

$\frac{\left[a+\frac{(35-1) d}{2}\right]}{\left[A+\frac{(35-1) D}{2}\right]}=\frac{5 * 35+4}{9 * 35+6}$

$\frac{a+17 d}{A+17 D}=\frac{179}{321}$

So the ratio of $18^{\text {th }}$ term of both the AP is 179:321