# How do you find the 7th term of 2(sqrt{5}), 4(sqrt{5}), 6(sqrt{5}), ...?

##### 1 Answer
Aug 31, 2016

$14 \sqrt{5}$.

#### Explanation:

Let the given seq. be ${\left\{{t}_{n}\right\}}_{1}^{\infty} = \left\{2 \sqrt{5} , 4 \sqrt{5} , 6 \sqrt{5} , \ldots\right\}$.

Then, ${t}_{2} - {t}_{1} = 2 \sqrt{5} , {t}_{3} - {t}_{2} = 2 \sqrt{5} , \ldots$

Assuming that the pattern continues , we find that the seq. is an A.P.,

with, ${t}_{1} = 2 \sqrt{5} \text{, and, common difference } d = 2 \sqrt{5}$.

$\therefore {t}_{n} = {t}_{1} + \left(n - 1\right) d = 2 \sqrt{5} + \left(2 \sqrt{5}\right) \left(n - 1\right) = \left(2 \sqrt{5}\right) \left(1 + n - 1\right) = \left(2 \sqrt{5}\right) n$.

Hence, the reqd. term$= {t}_{7} = \left(2 \sqrt{5}\right) 7 = 14 \sqrt{5}$.