Question

Find the sums given below :
(i) $7+10\frac{1}{2}+14+...+84$
(ii) $34+32+30+...+10$
(iii) $-5+(-8)+(-11)+...+(-230)$

Answer 1

(i) $7,10\frac{1}{2},\mathrm{14........84}$

Here, first $a=7$ and $l=84$

$d=10\frac{1}{2}-7=3\frac{1}{2}=\frac{7}{2}$

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

$84=7+(n-1)\frac{7}{2}\Rightarrow 84-7=\frac{7}{2}(n-1)\Rightarrow n-1=22\Rightarrow n=23$

and last term $l=84$

We know that $S_n=\frac{n}{2}(a+l)$

$S_n=\frac{23}{2}(7+84)$

$=\frac{23\times 91}{2}$

$=\frac{2093}{2}$

$=1046\frac{1}{2}$

(ii) $34,32,\mathrm{30....................10}$

Here $a=34,l=10$

$d=32-34=-2$

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

$10=34+(n-1)(-2)\Rightarrow 10-34=(-2)(n-1)\Rightarrow n-1=12\Rightarrow n=13$

and last term $l=10$

We know that $S_n=\frac{n}{2}(a+l)$

$S_n=\frac{13}{2}(34+10)$

$=\frac{13\times 44}{2}$

$=286$

(iii) $-5+(-8)+(-11)+...........+(-230)$

Here $a=-5,l=-230$

$d=(-8-(-5\left)\right)=-3$

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

$230=-5+(n-1)(-3)\Rightarrow -230+5=(-3)(n-1)\Rightarrow n-1=75\Rightarrow n=76$

and last term $l=-230$

We know that $S_n=\frac{n}{2}(a+l)$

$S_n=\frac{76}{2}(-5+(-230)$

$=-8930$