Question

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

• A. $-441$
• B. $441$
• C. $434$
• D. $-434$

Answer 1

Answer: (A)

$-441$

$18+15\frac{1}{2}+13+-----(-49\frac{1}{2})$

In A.P

$a_n=a+(n-1).d$

$a_n=-49\frac{1}{2},a=18,d=\frac{-5}{2}$

$18+(n-1)\left(\frac{-5}{2}\right)=\frac{-99}{2}$

$18+\frac{(-5n+5)}{2}=\frac{-99}{2}$

$18+\frac{(-5n+5)}{2}=\frac{-99}{2}$

$\frac{36-5n+5}{2}=\frac{-99}{2}$

$-5n=-99-41$

$n=28$

Now, we know that the sum of A.P is

$S_n=\frac{n}{2}[2a+(n-1).d]$

$=\frac{28}{2}[2\times 18+(28-1).(-\frac{5}{2}\left)\right]$

$=14[36-\frac{135}{2}]$

$=14\left[\frac{72-135}{2}\right]$

$=7[72-135]$

$=7(-63)$

$S_n=-441$

Hence the sum of above series is $-441$