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) $3,\frac{9}{2},6,\frac{15}{2},\dots \dots 25terms$
(iv) 41, 36, 31, ..... to 12 terms
(v) a + b, a - b, a - 3b, ... to 22 terms
(vi) $(x-y{)}^2,(x^2+y^2),(x+y{)}^2,\dots \dots \text{to n terms}$
(vii) $\frac{x-y}{x+y},\frac{3x-2y}{x+y},\frac{5x-3y}{x+y},\dots \dots \text{to n terms}$

Answer 1

(i) 50, 46, 42, ..... 10 terms
a = 50, d = (46 - 50) = - 4
n = 10
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]S_{10}=\frac{10}{2}[2\times 50+(10-1\left)\right(-4\left)\right]=320$

(ii) 1, 3, 5, 7, ..... 12 terms
$S_{12}=\frac{12}{2}[2\times 1+(12-1\left)2\right]=6\times 24=144$

(iii) $3,\frac{9}{2},6,\frac{15}{2},\dots \dots \text{25 terms}{S}_n=\frac{n}{2}[2a+(n-1\left)d\right]S_{25}=\frac{25}{2}(2\times 3+24\times \frac{3}{2})=525$

(iv) $41,36,31,\dots \dots \text{12 terms}{S}_n=\frac{n}{2}[2a+(n-1\left)d\right]S_{25}=\frac{12}{2}[2\times 41+(11\left)\right(-5\left)\right]=162$

(v) $a+b,a-b,a-3b,\dots \dots \text{to 22 terms}{S}_n=\frac{n}{2}[2a+(n-1\left)d\right]S_{22}=\frac{22}{2}[2a+2b+21(-2b\left)\right]=22a-440b$

(vi) $(x-y{)}^2,(x^2+y^2),(x+y{)}^2,\dots \dots \text{n terms}{S}_n=\frac{n}{2}[2a+(n-1\left)d\right]=\frac{n}{2}\left[2\right(x^2+y^2-2xy)+(x-1\left)\right(-2xy\left)\right]=n\left[\right(x-y{)}^2+(n-1)xy]$

(vii) $\frac{x-y}{x+y},\frac{3x-2y}{x+y},\frac{5x-3y}{x+y},\dots \dots \text{to n terms}$
nth term in above sequence is $\frac{(2n-1)x-ny}{x+y}$
Sum of n terms is given by
$\frac{1}{x+y}[x+3x+5x+\dots \dots +(2n-1)x-(y+2y+3y\dots \dots +ny\left)\right]=\frac{1}{x+y}\left[\frac{n}{2}\right(2x+(n-1)2x)-\frac{n(n+1)y}{2}]=\frac{1}{2}(x+y)[2n^{2}x-2n^{2}y-ny]=\frac{n}{2(x+y)}[2nx-2ny-y]=\frac{n}{2(x+y)}\left[n\right(2x-2y)-y]$