Question

The sum of the series $\frac { 1 } { 2 } + \frac { 1 } { 3 } + \frac { 1 } { 6 } + \ldots \ldots$ to 9 terms is.

• A. $- \frac { 5 } { 6 }$
• B. $- \frac { 1 } { 2 }$
• C. 1
• D. $- \frac { 3 } { 2 }$

Answer 1

Answer: (D)

$- \frac { 3 } { 2 }$

Answer 2

Given series $\frac { 1 } { 2 } + \frac { 1 } { 3 } + \frac { 1 } { 6 } + \ldots \ldots$

Here $a = \frac { 1 } { 2 }$ , common difference $d = - \frac { 1 } { 6 }$ and $n = 9$ .

$\Rightarrow$ $S _ { 9 } = \frac { 9 } { 2 } \left[ 2 \times \frac { 1 } { 2 } + ( 9 - 1 ) \left( - \frac { 1 } { 6 } \right) \right] = - \frac { 3 } { 2 }$ .