Find the sum:$18 + 15\frac{1}{2} + 13 + ……… + (-49\frac{1}{2})$

Given:

Given sequence is $18 + 15\frac{1}{2} + 13 + ……… + (-49\frac{1}{2})$.

To do:

We have to find the sum of $18 + 15\frac{1}{2} + 13 + ……… + (-49\frac{1}{2})$.

Solution:

Here,

$18 + 15\frac{1}{2} + 13 + ……… + (-49\frac{1}{2})$ is in A.P.

$a=18, d=15\frac{1}{2}-18=\frac{15(2)+1-18(2)}{2}=\frac{-5}{2}$ and \( l=-49\frac{1}{2} \)

We know that,

\( a_{n}=a+(n-1) d \)
\( \Rightarrow -49\frac{1}{2}=18+(n-1) \times \frac{-5}{2} \)
$\Rightarrow \frac{-49(2)-1}{2}=\frac{18(2)+(n-1)(-5)}{2}$

\( \Rightarrow -98-1=36-5n+5 \)
\( \Rightarrow 5n=41+99 \)

\( \Rightarrow n=\frac{140}{5}=28 \)
\( \mathrm{S}_{n}=\frac{n}{2}[a+l] \)

$=\frac{28}{2}[18+(-49\frac{1}{2})]$
$=14 \times\frac{18(2)-49(2)-1}{2}$
\( = 14 \times \frac{36-98-1}{2} \) 

\( = 14 \times \frac{-63}{2} \) 

\( = 7 \times (-63) \)

$= -441$ 

Therefore, the sum of the given sequence is $-441$.  

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Updated on: 10-Oct-2022

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