Question
Find the sum:
$18+15\frac{1}{2}+13\ + ..... \ +\Big(-49\frac{1}{2}\Big)$
$18+15\frac{1}{2}+13\ + ..... \ +\Big(-49\frac{1}{2}\Big)$
✓
Answer
$\text{S}=18+15\frac{1}{2}+13\ + ..... \ +\Big(-49\frac{1}{2}\Big)$Here, a = 18, $\text{l}=\text{a}_\text{n}=-49\frac{1}{2}=\frac{-99}{2}$
$\text{d}=\frac{31}{2}-18=\frac{-5}{2}$ For n $\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}$ $\frac{-99}{2}=18+(\text{n}-1)\Big(\frac{-5}{2}\Big)$ $\frac{-99}{2}=\frac{36-(\text{n}-1)5}{2}$ $-99=36-(\text{n}-1)5$ $(\text{n}-1)5=36+99$ $(\text{n}-1)5=135$ $\text{n}-1=\frac{135}{5}$ $\text{n}-1=27$ $\text{n}=27+1$ $\text{n}=28$ Now, $\text{S}_{28}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$ $=\frac{28}{2}\Big[2\times18+(28-1)\times\Big(\frac{-5}{2}\Big)\Big]$ $=14\Big[36+27\times\Big(\frac{-5}{2}\Big)\Big]$ $=14\Big[36-\frac{135}{2}\Big]$ $=14\Big[\frac{72-135}{2}\Big]=7\times(-63)=-441$
$\text{d}=\frac{31}{2}-18=\frac{-5}{2}$ For n $\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}$ $\frac{-99}{2}=18+(\text{n}-1)\Big(\frac{-5}{2}\Big)$ $\frac{-99}{2}=\frac{36-(\text{n}-1)5}{2}$ $-99=36-(\text{n}-1)5$ $(\text{n}-1)5=36+99$ $(\text{n}-1)5=135$ $\text{n}-1=\frac{135}{5}$ $\text{n}-1=27$ $\text{n}=27+1$ $\text{n}=28$ Now, $\text{S}_{28}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$ $=\frac{28}{2}\Big[2\times18+(28-1)\times\Big(\frac{-5}{2}\Big)\Big]$ $=14\Big[36+27\times\Big(\frac{-5}{2}\Big)\Big]$ $=14\Big[36-\frac{135}{2}\Big]$ $=14\Big[\frac{72-135}{2}\Big]=7\times(-63)=-441$
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