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)
(iv) 5 + (–41) + 9 + (–39) + 13 + (–37) + 17 + ... + (–5) + 81 + (–3)

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 = $\frac{76}{2}\times \left[\left(-5\right)+\left(-230\right)\right]$ $\left[S_n=\frac{n}{2}\left(a+l\right)\right]$
$$\begin{gathered} =\frac{76}{2}\times \left(-235\right) \\ =-8930 \end{gathered}$$
(iv)
$$\begin{gathered} \mathrm{Given}\mathrm{series}\mathrm{is}: \\ 5+\left(-41\right)+9+\left(-39\right)+13+\left(-37\right)+17+...+\left(-5\right)+81+\left(-3\right) \\ \mathrm{We}\mathrm{have}\mathrm{two}\mathrm{series} \\ 1\mathrm{st}:5+9+13+17+...+81 \\ 2\mathrm{nd}:\left(-41\right)+\left(-39\right)+\left(-37\right)+...+\left(-3\right) \\ \mathrm{Let}\mathrm{us}\mathrm{consider}1\mathrm{st}\mathrm{series}\mathrm{first}: \\ a=5 \\ d=9-5=4 \\ n\mathrm{th}\mathrm{term}=81 \\ {a}_n=a+\left(n-1\right)d \\ 81=5+\left(n-1\right)4 \\ n=20 \end{gathered}$$
Using formula of sum
$$\begin{gathered} S_n=\frac{n}{2}\left(a+a_n\right) \\ {S}_n=\frac{20}{2}\left(5+81\right) \\ {S}_n=860 \\ \mathrm{Now},\mathrm{consider}2\mathrm{nd}\mathrm{series}: \\ \left(-41\right)+\left(-39\right)+\left(-37\right)+...+\left(-3\right) \\ a=-41 \\ d=-39+41=2 \\ n\mathrm{th}\mathrm{term}=-3 \\ a+\left(n-1\right)d=-3 \\ -41+\left(n-1\right)2=-3 \\ \Rightarrow n=20 \end{gathered}$$
Using formula of sum
$$\begin{gathered} s_n=\frac{n}{2}\left(a+a_n\right) \\ {s}_n=\frac{20}{2}\left(-41+\left(-3\right)\right) \\ {s}_n=\frac{20}{2}\left(-44\right) \\ {s}_n=-440 \\ \mathrm{Sum}\mathrm{of}\mathrm{total}\mathrm{series}=\mathrm{sum}\mathrm{of}1\mathrm{st}+\mathrm{sum}\mathrm{of}2\mathrm{nd}\mathrm{series} \\ \mathrm{Sum}\mathrm{of}\mathrm{given}\mathrm{arithmetic}\mathrm{series}=860-440=420 \end{gathered}$$