Question

Find the sum:  \[18 + 15\frac{1}{2} + 13 + . . . + \left( - 49\frac{1}{2} \right)\]

 

Answer 1

\[18 + 15\frac{1}{2} + 13 + . . . + \left( - 49\frac{1}{2} \right)\]

Common difference of the A.P. (d) =  a₂ - a₁ 

 \[= 15\frac{1}{2} - 18\]
\[ = \frac{31}{2} - 18\]
\[ = \frac{31 - 36}{2}\]
\[ = \frac{- 5}{2}\]

So here,

First term (a) = 18

Last term (l) = \[- 49\frac{1}{2} = \frac{- 99}{2}\] 

Common difference (d) = \[\frac{- 5}{2}\]

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

`a_n = a+(n-1)d`

So, for the last term,

\[\frac{- 99}{2} = 18 + \left( n - 1 \right)\frac{- 5}{2}\]
\[\frac{- 99}{2} = 18 + \left( \frac{- 5}{2} \right)n + \frac{5}{2}\]
\[\frac{5}{2}n = 18 + \frac{5}{2} + \frac{99}{2}\]
\[\frac{5}{2}n = 18 + \frac{104}{2}\]
\[n = 28\]

Now, using the formula for the sum of n terms, we get

\[S_n = \frac{28}{2}\left[ 2 \times 18 + \left( 28 - 1 \right)\left( \frac{- 5}{2} \right) \right]\]
\[ S_n = 14\left[ 36 + 27\left( \frac{- 5}{2} \right) \right]\]
\[ S_n = - 441\]

Therefore, the sum of the A.P is   \[S_n = - 441\]