Question

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

Answer 1

`14sqrt5`.

Explanation:

Let the given seq. be `{t_n}_1^oo ={2sqrt5,4sqrt5,6sqrt5,...}`.

Then, `t_2-t_1=2sqrt5, t_3-t_2=2sqrt5,...`

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

with, `t_1=2sqrt5", and, common difference "d=2sqrt5`.

`:. t_n=t_1+(n-1)d=2sqrt5+(2sqrt5)(n-1)=(2sqrt5)(1+n-1)=(2sqrt5)n`.

Hence, the reqd. term`=t_7=(2sqrt5)7=14sqrt5`.