Question

Then the sum of elements of 10^th row. 

(1) 1505 

(2) 1438 

(3) 1981 

(4) 1745

Answer 1

Correct option is (1) 1505 

The given sequence is 

2, (5, 8), (11, 14, 17), (20, 21, 24, 27) … 

Before 10^th term total number of numbers would be 

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 9 x 5 = 45 

It is an A.P. with a = 2 and d = 3 

T₄₆ = 2 + 45 x 3 = 137

Sum of 10^th terms = 10/2 (2 x 137 + (10 – 1)3) = 1505