If the ratio of the sum of the first n terms of two A.Ps is (7n + 1) : (4n + 27)

Arithmetic Progressions (10)

If the ratio of the sum of the first n terms of two A.Ps is (7n + 1) : (4n + 27), then find the ratio of their 9th terms.

Answer

Let the first terms be a and a' and d and d' be their respective common differences.

$$ \frac{S_n}{S'_n} = \frac{\frac{n}{2}(2a + (n -1)d)}{\frac{n}{2}{(2a' + (n-1) d')}} = \frac{7n +1}{4n+27} $$

$$ a = \frac{a + \left( \frac{n-1}{2}\right)d}{a +\left( \frac{n-1}{2}\right)d')} = \frac{7n +1}{4n +27}$$

$$  \text {To get ratio of 9th terms}, replacing \frac{n-1}{2} = 8 $$

$$ n = 17 $$

$$ \text{Hence} \frac{t_9}{t'_9} = \frac{a + 8d}{a' + 8d'} = \frac{120}{95} \text{or} \frac{24}{19} $$

Exam Year: 2017