Find the sum of the following arithmetic progressions:

Solution:

In the given problem, we need to find the sum of terms for different arithmetic progressions. So, here we use the following formula for the sum of n terms of an A.P.,

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

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

(i)  To 10 terms

Common difference of the A.P. (d)

$=a_{2}-a_{1}$

$=46-50$

$=-4$

Number of terms (n) = 10

First term for the given A.P. (a) = 50

So, using the formula we get,

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

$=(5)[100+(9)(-4)]$

$=(5)[100-36]$

$=(5)[64]$

 

$=320$

Therefore, the sum of first 10 terms for the given A.P. is.

(ii) $1,3,5,7, \ldots-26$ To 12 terms.

Common difference of the A.P. (d)

$=a_{2}-a_{1}$

$=3-1$

 

$=2$

Number of terms (n) = 12

First term for the given A.P. (a) = 1

So, using the formula we get,

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

$=(6)[2+(11)(2)]$

$=(6)[2+22]$

$=(6)[24]$

$=144$

Therefore, the sum of first 12 terms for the given A.P. is.

(iii) $3,9 / 2,6,15 / 2, \ldots$ To 25 terms.

Common difference of the A.P. $(d)=a_{2}-a_{1}$

$=\frac{9}{2}-3$

$=\frac{9-6}{2}$

 

$=\frac{3}{2}$

Number of terms (n) = 25

First term for the given A.P. (a) = 3

So, using the formula we get,

$S_{25}=\frac{25}{2}\left[2(3)+(25-1)\left(\frac{3}{2}\right)\right]$

$=\left(\frac{25}{2}\right)\left[6+(24)\left(\frac{3}{2}\right)\right]$

$=\left(\frac{25}{2}\right)\left[6+\left(\frac{72}{2}\right)\right]$

$=\left(\frac{25}{2}\right)[6+36]$

$=\left(\frac{25}{2}\right)[42]$

$=(25)(21)$

$=525$

On further simplifying, we get,

$S_{25}=525$

Therefore, the sum of first 25 terms for the given A.P. is.

(iv) $41,36,31, \ldots$ To 12 terms.

Common difference of the A.P. (d) = 

$=36-41$

 

$=-5$

Number of terms (n) = 12

First term for the given A.P. (a) = 41

So, using the formula we get,

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

$=(6)[82+(11)(-5)]$

$=(6)[82-55]$

$=(6)[27]$

 

$=162$

Therefore, the sum of first 12 terms for the given A.P. is 162 .

(v) $a+b, a-b, a-3 b, \ldots$ To 22 terms.

Common difference of the A.P. $(d)=a_{2}-a_{1}$

$=(a-b)-(a+b)$

$=a-b-a-b$

 

$=-2 b$

Number of terms (n) = 22

First term for the given A.P. (a) = 

So, using the formula we get,

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

$=(11)[2 a+2 b+(21)(-2 b)]$

$=(11)[2 a+2 b-42 b]$

$=(11)[2 a-40 b]$

 

$=22 a-440 b$

Therefore, the sum of first 22 terms for the given A.P. is $22 a-440 b$.

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

Common difference of the A.P. $(d)=a_{2}-a_{1}$

$=\left(x^{2}+y^{2}\right)-(x-y)^{2}$

$=x^{2}+y^{2}-\left(x^{2}+y^{2}-2 x y\right)$

$=x^{2}+y^{2}-x^{2}-y^{2}+2 x y$

 

$=2 x y$

Number of terms $(n)=n$

First term for the given A.P. $(a)=(x-y)^{2}$

 

So, using the formula we get,

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

Now, taking 2 common from both the terms inside the bracket we get,

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

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

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

Therefore, the sum of first $n$ terms for the given A.P. is $n\left[(x-y)^{2}+(n-1) x y\right]$

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

Number of terms (n) = n

First term for the given A.P. $(a)=\left(\frac{x-y}{x+y}\right)$

Common difference of the A.P. $(d)=a_{2}-a_{1}$

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

$=\frac{(3 x-2 y)-(x-y)}{x+y}$

$=\frac{3 x-2 y-x+y}{x+y}$

$=\frac{2 x-y}{x+y}$

So, using the formula we get,

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

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

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

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

Now, on further solving the above equation we get,

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

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

Therefore, the sum of first $n$ terms for the given A.P. is $\left(\frac{n}{2}\right)\left(\frac{n(2 x-y)-y}{x+y}\right)$.

(viii) $-26,-24,-22, \ldots$ To 36 terms.

Common difference of the A.P. $(d)=a_{2}-a_{1}$

$=(-24)-(-26)$

$=-24+26$

 

$=2$

Number of terms (n) = 36

First term for the given A.P. (a) = −26

So, using the formula we get,

$S_{36}=\frac{36}{2}[2(-26)+(36-1)(2)]$

$=(18)[-52+(35)(2)]$

$=(18)[-52+70]$

$=(18)[18]$

 

$=324$

Therefore, the sum of first 36 terms for the given A.P. is 324 .