Find the sum of each of the following arithmetic series:

(i) The given arithmetic series is $7+10 \frac{1}{2}+14+\ldots+84$.

Here, $a=7, d=10 \frac{1}{2}-7=\frac{21}{2}-7=\frac{21-14}{2}=\frac{7}{2}$ and $I=84$

Let the given series contain n terms. Then,

$a_{n}=84$

$\Rightarrow 7+(n-1) \times \frac{7}{2}=84 \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow \frac{7}{2} n+\frac{7}{2}=84$

$\Rightarrow \frac{7}{2} n=84-\frac{7}{2}=\frac{161}{2}$

$\Rightarrow n=\frac{161}{7}=23$

$\therefore$ Required sum $=\frac{23}{2} \times(7+84) \quad\left[S_{n}=\frac{n}{2}(a+l)\right]$

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

$=\frac{2093}{2}$

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

(ii) The given arithmetic series is 34 + 32 + 30 + ... + 10.

Here, a = 34, d = 32 − 34 = −2  and l = 10.

Let the given series contain n terms. Then,

$a_{n}=10$

$\Rightarrow 34+(n-1) \times(-2)=10 \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow-2 n+36=10$

$\Rightarrow-2 n=10-36=-26$

$\Rightarrow n=13$

$\therefore$ Required sum $=\frac{13}{2} \times(34+10) \quad\left[S_{n}=\frac{n}{2}(a+l)\right]$

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

$=286$

(iii) The given arithmetic series is (−5) + (−8) + (−11) + ... + (−230).

Here, a = −5, d = −8 − (−5) = −8 + 5 = −3  and l = −230.

Let the given series contain n terms. Then,

$a_{n}=-230$

$\Rightarrow-5+(n-1) \times(-3)=-230 \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow-3 n-2=-230$

$\Rightarrow-3 n=-230+2=-228$

$\Rightarrow n=76$

$\therefore$ Required sum $=\frac{76}{2} \times[(-5)+(-230)] \quad\left[S_{n}=\frac{n}{2}(a+l)\right]$

$=\frac{76}{2} \times(-235)$

$=-8930$

(iv)

Given series is :

$5+(-41)+9+(-39)+13+(-37)+17+\ldots+(-5)+81+(-3)$

We have two series

1st: $5+9+13+17+\ldots+81$

2nd: $(-41)+(-39)+(-37)+\ldots+(-3)$

Let us consider 1 st series first:

$a=5$

$d=9-5=4$

$n$th term $=81$

$a_{n}=a+(n-1) d$

$81=5+(n-1) 4$

$n=20$

Using formula of sum

$S_{n}=\frac{n}{2}\left(a+a_{n}\right)$

$S_{n}=\frac{20}{2}(5+81)$

$S_{n}=860$

Now, consider 2 nd series:

$(-41)+(-39)+(-37)+\ldots+(-3)$

$a=-41$

$d=-39+41=2$

$n$th term $=-3$

$a+(n-1) d=-3$

$-41+(n-1) 2=-3$

$\Rightarrow n=20$

Using formula of sum

$s_{n}=\frac{n}{2}\left(a+a_{n}\right)$

$s_{n}=\frac{20}{2}(-41+(-3))$

$s_{n}=\frac{20}{2}(-44)$

$s_{n}=-440$

Sum of total series $=$ sum of 1 st $+$ sum of 2 nd series

Sum of given arithmetic series $=860-440=420$