6 + \sqrt216;7+\sqrt343;8+\sqrt512;9 +\sqrt729; ?

SSC CHSL 15 Nov 2015 Shift 2 Paper

In question numbers 36 to 39, a series is given, with one term missing. Choose the correct alternative from the given ones that will complete the series.

Question : 38 of 200

Marks: +1, -0

6 +

√216;7+√343;8+√512;9

+√729;

10 + $√10000$
10 + $√10^5$
10 + $√100$
10 + $√1000$

Validate

Solution:

The pattern followed is =

n + √n^3

Eg :- 6 +

√216;7+√343;8+√512;9

+√729;

Now,

10 ^3 = 1000

Thus, the next term =

10 + √1000

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