# Find the sum of the following arithmetic progressions:

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 + ba − ba − 3b, ... to 22 terms

(vi) $(x-y)^{2},\left(x^{2}+y^{2}\right),(x+y)^{2}, \ldots$ to $n$ terms

(vii) $\frac{x-y}{x+y}, \frac{3 x-2 y}{x+y}, \frac{5 x-3 y}{x+y}, \ldots$ to $n$ terms

Solution:

(i) 50, 46, 42 ... to 10 terms

We have:

$a=50, d=(46-50)=-4$

$n=10$

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{10}{2}[2 \times 50+(10-1)(-4)]$

$=5[100-36]=320$

(ii) 1, 3, 5, 7 ... to 12 terms

We have:

$a=1, d=(3-1)=2$

$n=12$

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{12}{2}[2 \times 1+(12-1)(2)]$

$=6[24]=144$

(iii) 3, 9/2, 6, 15/2 ... to 25 terms

We have:

$a=3, d=(9 / 2-3)=3 / 2$

$n=25$

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{25}{2}[2 \times 3+(25-1)(3 / 2)]$

$=\frac{25}{2} \times 42$

$=525$

(iv) 41, 36, 31 ... to 12 terms

We have:

$a=41, d=(36-41)=-5$

$n=12$

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{12}{2}[2 \times 41+(12-1)(-5)]$

$=6 \times 27=162$

(v) a + ba − ba − 3b ... to 22 terms

We have:

First term $=a+b, d=(a-b-a-b)=-2 b$

$n=22$

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{22}{2}[2 \times(a+b)+(22-1)(-2 b)]$

$=11[2 a-40 b]=22 a-440 b$

(vi) (x − y)2, (x2 + y2), (x + y)2 ... to n terms

We have:

$a=(\mathrm{x}-\mathrm{y})^{2}, d=\left(x^{2}+y^{2}-(\mathrm{x}-\mathrm{y})^{2}\right)=2 x y$

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{n}{2}\left[2(\mathrm{x}-\mathrm{y})^{2}+(n-1)(2 x y)\right]$

$=\frac{n}{2} \times 2\left[(\mathrm{x}-\mathrm{y})^{2}+(n-1)(x y)\right]$

$=n\left[(\mathrm{x}-\mathrm{y})^{2}+(n-1)(x y)\right]$

(vii) $\frac{x-y}{x+y}, \frac{3 x-2 y}{x+y}, \frac{5 x-3 y}{x+y} \ldots$ to $n$ terms

We have:

$a=\frac{x-y}{x+y}, d=\left(\frac{3 x-2 y}{x+y}-\frac{x-y}{x+y}\right)=\left(\frac{2 x-y}{x+y}\right)$

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{n}{2}\left[2\left(\frac{x-y}{x+y}\right)+(n-1)\left(\frac{2 x-y}{x+y}\right)\right]$

$=\frac{n}{2(x+y)}[(2 x-2 y)+(2 x-y)(n-1)]$

$=\frac{n}{2(x+y)}[2 x-2 y-2 x+y+n(2 x-y)]$

$=\frac{n}{2(x+y)}[n(2 x-y)-y]$