Question

99th term of the series 2 + 7 + 14 + 23 + 34 + ............ is.

• A. $9998$
• B. $9999$
• C. $10000$
• D. $100000$

Answer 1

The correct option is A $9998$

Let $S_n=2+7+14+23+34+...........n$ terms
$\therefore S_{n-1}=2+7+14+23+34+...........n-1$ terms
Now $T_n=S_n-S_{n-1}=2+5+7+9+11+..................+n=(3+5+7+9+11+..................+n)-1$
$=\frac{n}{2}(6+(n-1\left)2\right)-1=n^2+2n-1$
Hence ${99}^{th}$ term is,
$T_{99}=99\times 101-1=9998$
Hence, option 'A' is correct.