Question

Find the sum of each of the following arithmetic series:

(i) $7+10\frac{1}{2}+14+...+84$
(ii) 34 + 32 + 30 + ... + 10
(iii) (−5) + (−8) + (−11) + ... + (−230)

Answer 1

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

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

Let the given series contain n terms. Then,

$$\begin{gathered} a_n=84 \\ \Rightarrow 7+\left(n-1\right)\times \frac{7}{2}=84\left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow \frac{7}{2}n+\frac{7}{2}=84 \\ \Rightarrow \frac{7}{2}n=84-\frac{7}{2}=\frac{161}{2} \end{gathered}$$
$\Rightarrow n=\frac{161}{7}=23$
∴ Required sum = $\frac{23}{2}\times \left(7+84\right)$ $\left[S_n=\frac{n}{2}\left(a+l\right)\right]$
$$\begin{gathered} =\frac{23}{2}\times 91 \\ =\frac{2093}{2} \\ =1046\frac{1}{2} \end{gathered}$$

(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,

$$\begin{gathered} a_n=10 \\ \Rightarrow 34+\left(n-1\right)\times \left(-2\right)=10\left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow -2n+36=10 \\ \Rightarrow -2n=10-36=-26 \end{gathered}$$
$\Rightarrow n=13$
∴ Required sum = $\frac{13}{2}\times \left(34+10\right)$ $\left[S_n=\frac{n}{2}\left(a+l\right)\right]$
$$\begin{gathered} =\frac{13}{2}\times 44 \\ =286 \end{gathered}$$

(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,

$$\begin{gathered} a_n=-230 \\ \Rightarrow -5+\left(n-1\right)\times \left(-3\right)=-230\left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow -3n-2=-230 \\ \Rightarrow -3n=-230+2=-228 \end{gathered}$$
$\Rightarrow n=76$
∴ Required sum = $\frac{76}{2}\times \left[\left(-5\right)+\left(-230\right)\right]$ $\left[S_n=\frac{n}{2}\left(a+l\right)\right]$
$$\begin{gathered} =\frac{76}{2}\times \left(-235\right) \\ =-8930 \end{gathered}$$