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

1 Answer
Aug 31, 2016

Answer:

#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#.