Question

Find the sum of each of the following APs:

(i) 2, 7, 12, 17, ... to 19 terms
(ii) 9, 7, 5, 3, ... to 14 terms
(iii) −37, −33, −29, ... to 12 terms
(iv) $\frac{1}{15},\frac{1}{12},\frac{1}{10},...$ to 11 terms
(v) 0.6, 1.7, 2.8, ... to 100 terms

Answer 1

(i) The given AP is 2, 7, 12, 17, ... .

Here, a = 2 and d = 7 − 2 = 5

Using the formula, $S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we have
$$\begin{gathered} S_{19}=\frac{19}{2}\left[2\times 2+\left(19-1\right)\times 5\right] \\ =\frac{19}{2}\times \left(4+90\right) \\ =\frac{19}{2}\times 94 \\ =893 \end{gathered}$$

(ii) The given AP is 9, 7, 5, 3, ... .

Here, a = 9 and d = 7 − 9 = −2

Using the formula, $S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we have
$$\begin{gathered} S_{14}=\frac{14}{2}\left[2\times 9+\left(14-1\right)\times \left(-2\right)\right] \\ =7\times \left(18-26\right) \\ =7\times \left(-8\right) \\ =-56 \end{gathered}$$

(iii) The given AP is −37, −33, −29, ... .

Here, a = −37 and d = −33 − (−37) = −33 + 37 = 4

Using the formula, $S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we have
$$\begin{gathered} S_{12}=\frac{12}{2}\left[2\times \left(-37\right)+\left(12-1\right)\times 4\right] \\ =6\times \left(-74+44\right) \\ =6\times \left(-30\right) \\ =-180 \end{gathered}$$

(iv) The given AP is $\frac{1}{15},\frac{1}{12},\frac{1}{10},...$ .

Here, a = $\frac{1}{15}$ and d = $\frac{1}{12}-\frac{1}{15}=\frac{5-4}{60}=\frac{1}{60}$

Using the formula, $S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we have
$$\begin{gathered} S_{11}=\frac{11}{2}\left[2\times \left(\frac{1}{15}\right)+\left(11-1\right)\times \frac{1}{60}\right] \\ =\frac{11}{2}\times \left(\frac{2}{15}+\frac{10}{60}\right) \\ =\frac{11}{2}\times \left(\frac{18}{60}\right) \\ =\frac{33}{20} \end{gathered}$$

(v) The given AP is 0.6, 1.7, 2.8, ... .

Here, a = 0.6 and d = 1.7 − 0.6 = 1.1

Using the formula, $S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we have
$$\begin{gathered} S_{100}=\frac{100}{2}\left[2\times 0.6+\left(100-1\right)\times 1.1\right] \\ =50\times \left(1.2+108.9\right) \\ =50\times 110.1 \\ =5505 \end{gathered}$$