Question

Find the sum of the following arithmetic progressions:
(i) 50, 46, 42, ... to 10 terms
(ii) 1, 3, 5, 7, ... to 12 terms
(iii) 3, 9/2, 6, 15/2, ... to 25 terms
(iv) 41, 36, 31, ... to 12 terms
(v) a + b, a − b, a − 3b, ... to 22 terms
(vi) (x − y)², (x² + y²), (x + y)², ... to n terms
(vii) $\frac{x-y}{x+y},\frac{3x-2y}{x+y},\frac{5x-3y}{x+y}$, ... to n terms

Answer 1

(i) 50, 46, 42 ... to 10 terms
$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ a=50,d=\left(46-50\right)=-4 \\ n=10 \\ {S}_n=\frac{n}{2}\left[2a+(n-1)d\right] \\ =\frac{10}{2}\left[2\times 50+(10-1)(-4)\right] \\ =5\left[100-36\right]=320 \end{gathered}$$

(ii) 1, 3, 5, 7 ... to 12 terms
$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ a=1,d=\left(3-1\right)=2 \\ n=12 \\ {S}_n=\frac{n}{2}\left[2a+(n-1)d\right] \\ =\frac{12}{2}\left[2\times 1+(12-1)\left(2\right)\right] \\ =6\left[24\right]=144 \end{gathered}$$

(iii) 3, 9/2, 6, 15/2 ... to 25 terms
$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ a=3,d=\left(9/2-3\right)=3/2 \\ n=25 \\ {S}_n=\frac{n}{2}\left[2a+(n-1)d\right] \\ =\frac{25}{2}\left[2\times 3+(25-1)(3/2)\right] \\ =\frac{25}{2}\times 42 \\ =525 \end{gathered}$$

(iv) 41, 36, 31 ... to 12 terms
$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ a=41,d=\left(36-41\right)=-5 \\ n=12 \\ {S}_n=\frac{n}{2}\left[2a+(n-1)d\right] \\ =\frac{12}{2}\left[2\times 41+(12-1)(-5)\right] \\ =6\times 27=162 \end{gathered}$$

(v) a + b, a − b, a − 3b ... to 22 terms
$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ \mathrm{First}\mathrm{term}=a+b,d=\left(a-b-a-b\right)=-2b \\ n=22 \\ {S}_n=\frac{n}{2}\left[2a+(n-1)d\right] \\ =\frac{22}{2}\left[2\times (a+b)+(22-1)(-2b)\right] \\ =11\left[2a-40b\right]=22a-440b \end{gathered}$$

(vi) (x − y)², (x² + y²), (x + y)² ... to n terms
$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ a={(x\; -\; y)}^2,d=\left(x^2+y^2-{(x\; -\; y)}^2\right)=2xy \\ {S}_n=\frac{n}{2}\left[2a+(n-1)d\right] \\ =\frac{n}{2}\left[2{(x\; -\; y)}^2+(n-1)\left(2xy\right)\right] \\ =\frac{n}{2}\times 2\left[{(x\; -\; y)}^2+(n-1)\left(xy\right)\right] \\ =n\left[{(x\; -\; y)}^2+(n-1)\left(xy\right)\right] \end{gathered}$$

(vii) $\frac{x-y}{x+y},\frac{3x-2y}{x+y},\frac{5x-3y}{x+y}$ ... to n terms
$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ a=\frac{x-y}{x+y},d=\left({\displaystyle \frac{3x-2y}{x+y}}-{\displaystyle \frac{x-y}{x+y}}\right)=\left(\frac{2x-y}{x+y}\right) \\ {S}_n=\frac{n}{2}\left[2a+(n-1)d\right] \\ =\frac{n}{2}\left[2\left(\frac{x-y}{x+y}\right)+(n-1)\left(\frac{2x-y}{x+y}\right)\right] \\ =\frac{n}{2(x+y)}\left[(2x-2y)+(2x-y)(n-1)\right] \\ =\frac{n}{2(x+y)}\left[2x-2y-2x+y+n(2x-y)\right] \\ =\frac{n}{2(x+y)}\left[n(2x-y)-y\right] \end{gathered}$$