Question

If the natural numbers are divided into groups of {1}, {2, 3}, {4, 5, 6}, {7, 8, 9, 10} ....Then the /sum of 50th group is

• A. 65225
• B. 56225
• C. 62525
• D. 53625

Answer 1

The correct option is C 62525

Pattern:

1

2 3

4 5 6

7 8 9 10

..........

Series for starting number:

1, 2, 4, 7, ...

Let $T_n=a{n}^2+bn+c$

a+b+c=1 ......(1)

4a+2b+c=2 ......(2)

9a+3b+c=4 .......(3)

From eq (1), (2), (3), we get:

$a=\frac{1}{2},b=-\frac{1}{2},c=1$

$\therefore T_n=\frac{n^2}{2}-\frac{n}{2}+1$

$\therefore T_{50}=\frac{{50}^2}{2}-\frac{50}{2}+1=1226$

Now, elements in 50th group:

1226, 1227, 1228,.....(50 terms)

$\therefore S_n=\frac{50}{2}[2\times 1226+(50-1)\times 1]=62525$