Find the sums given below :

Question.

Find the sums given below :

(i) $7+10 \frac{1}{2}+14+\ldots+84$

(ii) 34 + 32 + 30 + ... + 10

(iii) – 5 + (– 8) + (– 11) +...+ (– 230).

Solution:

(i) $\mathrm{a}=7, \mathrm{~d}=10 \frac{1}{2}-7=3 \frac{1}{2}=\frac{7}{2}$

$\ell=t_{n}=84 \Rightarrow a+(n-1) d=84$

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

$\Rightarrow(\mathrm{n}-1) \times \quad \frac{7}{2}=77$

$\Rightarrow \mathrm{n}-1=77 \times \frac{2}{7}=22$

$\Rightarrow \mathrm{n}=23$

The sum $=\frac{n}{2}\left\{a+t_{n}\right)=\frac{23}{2}\{7+84)$

$=\frac{23}{2} \times 91=\frac{2093}{2}=1046 \frac{1}{2}$

(ii) $34+32+30+\ldots \ldots .+10$

$a=34$

$d=a_{2}-a_{1}=32-34=-2$

$\ell=10$

Let 10 be the nth term of this A.P.

$\ell=a+(n-1) d$

$10=34+(\mathrm{n}-1)(-2)$

$-24=(n-1)(-2)$

$12=n-1$

$n=13$

$S_{n}=\frac{n}{2}(a+\ell)$

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

$=\frac{13 \times 44}{2}=13 \times 22=286$

(iii) $(-5)+(-8)+(-11)+\ldots \ldots \ldots+(-230)$

For this A.P.,

$a=-5$

$\ell=-230$

$d=a_{2}-a_{1}=(-8)-(-5)$

$=-8+5=-3$

Let $-230$ be the $n^{\text {th }}$ term of this A.P.

$\ell=a+(n-1) d$

$-230=-5+(n-1)(-3)$

$-225=(n-1)(-3)$

$(n-1)=75$

$n=76$

And, $S_{n}=\frac{n}{2}(a+\ell)$

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

$=38(-235)$

$=-8930$