Find the sum:

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) $2+4+6+\ldots+200$

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

$=6-4$

 

$=2$

So here,

First term (a) = 2

Last term (l) = 200

Common difference (d) = 2

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

$a_{e}=a+(n-1) d$

So, for the last term,

$200=2+(n-1) 2$

$200=2+2 n-2$

 

$200=2 n$

Further simplifying,

$n=\frac{200}{2}$

$n=100$

Now, using the formula for the sum of n terms, we get

$S_{n}=\frac{100}{2}[2(2)+(100-1) 2]$

$=50[4+(99) 2]$

$=50(4+198)$

On further simplification, we get,

$S_{n}=50(202)$

$=10100$

Therefore, the sum of the A.P is $S_{n}=10100$

(ii) $3+11+19+\ldots+803$

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

$=19-11$

 

$=8$

So here,

First term (a) = 3

Last term (l) = 803

Common difference (d) = 8

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

$a_{n}=a+(n-1) d$

So, for the last term,

Further simplifying,

$803=3+(n-1) 8$

$803=3+8 n-8$

$803+5=8 n$

$808=8 n$

$n=\frac{808}{8}$

$n=101$

Now, using the formula for the sum of n terms, we get

$S_{\pi}=\frac{101}{2}[2(3)+(101-1) 8]$

$=\frac{101}{2}[6+(100) 8]$

$=\frac{101}{2}(806)$

$=101(403)$

 

$=40703$

Therefore, the sum of the A.P is $S_{n}=40703$

(iii) $(-5)+(-8)+(-11)+\ldots+(-230)$

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

$=-8-(-5)$

$=-8+5$

 

$=-3$

So here,

First term (a) = −5

Last term (l) = −230

Common difference (d) = −3

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

$a_{n}=a+(n-1) d$

So, for the last term,

$-230=-5+(n-1)(-3)$

$-230=-5-3 n+3$

$-230+2=-3 n$

$\frac{-228}{-3}=n$

$n=76$

Now, using the formula for the sum of n terms, we get

$S_{n}=\frac{76}{2}[2(-5)+(76-1)(-3)]$

$=38[-10+(75)(-3)]$

$=38(-10-225)$

$=38(-235)$

$=-8930$

Therefore, the sum of the A.P is $S_{n}=-8930$]

(iv) $1+3+5+7 \ldots+199$

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

$=3-1$

$=2$

So here,

First term (a) = 1

Last term (l) = 199

Common difference (d) = 2

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

$a_{n}=a+(n-1) d$

So, for the last term,

$199=1+(n-1) 2$

 

$199=1+2 n-2$

$199+1=2 n$

$n=\frac{200}{2}$

 

$n=100$

Now, using the formula for the sum of n terms, we get

$S_{n}=\frac{100}{2}[2(1)+(100-1) 2]$

$=50[2+(99) 2]$

 

$=50(2+198)$

On further simplification, we get,

$S_{n}=50(200)$

$=10000$

Therefore, the sum of the A.P is $S_{n}=10000$

(v) $7+10 \frac{1}{2}+14+\ldots+84$

Common difference of the A.P is

$(d)=$

$=10 \frac{1}{2}-7$

$=\frac{21}{2}-7$

$=\frac{21-14}{2}$

 

$=\frac{7}{2}$

So here,

First term (a) = 7

Last term (l) = 84

Common difference $(d)=\frac{7}{2}$

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

$a_{n}=a+(n-1) d$

So, for the last term,

$84=7+(n-1) \frac{7}{2}$

$84=7+\frac{7 n}{2}-\frac{7}{2}$

$84=\frac{14-7}{2}+\frac{7 n}{2}$

$84(2)=7+7 n$

Further solving for n,

$7 n=168-7$

$n=\frac{161}{7}$

$n=23$

Now, using the formula for the sum of n terms, we get

$S_{n}=\frac{23}{2}\left[2(7)+(23-1) \frac{7}{2}\right]$

$=\frac{23}{2}\left[14+(22) \frac{7}{2}\right]$

$=\frac{23}{2}(14+77)$

$=\frac{23}{2}(91)$

On further simplification, we get,

$S_{n}=\frac{2093}{2}$

Therefore, the sum of the A.P is $S_{n}=\frac{2093}{2}$

(vi) $34+32+30+\ldots+10$

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

$=32-34$

 

$=-2$

So here,

First term (a) = 34

Last term (l) = 10

Common difference (d) = −2

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

$a_{n}=a+(n-1) d$

So, for the last term,

$10=34+(n-1)(-2)$

$10=34-2 n+2$

 

$10=36-2 n$

$10-36=-2 n$

Further solving for n,

$-2 n=-26$

$n=\frac{-26}{-2}$

$n=13$

 

Now, using the formula for the sum of n terms, we get

$S_{n}=\frac{13}{2}[2(34)+(13-1)(-2)]$

$=\frac{13}{2}[68+(12)(-2)]$

$=\frac{13}{2}(68-24)$

$=\frac{13}{2}(44)$

On further simplification, we get,

$S_{n}=13(22)$

$=286$

Therefore, the sum of the A.P is $S_{n}=286$

(vii) $25+28+31+\ldots+100$

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

$=28-25$

 

$=3$

So here,

First term (a) = 25

Last term (l) = 100

Common difference (d) = 3

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

$a_{n}=a+(n-1) d$

So, for the last term,

$100=25+(n-1)(3)$

$100=25+3 n-3$

 

$100=22+3 n$

$100-22=3 n$

Further solving for n,

$78=3 n$

$n=\frac{78}{3}$

$n=26$

Now, using the formula for the sum of n terms, we get

$S_{n}=\frac{26}{2}[2(25)+(26-1)(3)]$

$=13[50+(25)(3)]$

$=13(50+75)$

 

$=13(125)$

On further simplification, we get,

$S_{n}=1625$

Therefore, the sum of the A.P is $S_{n}=1625$.

(viii) $18+15 \frac{1}{2}+13+\ldots+\left(-49 \frac{1}{2}\right)$

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

$==15 \frac{1}{2}-18$

$=\frac{31}{2}-18$

$=\frac{31-36}{2}$

 

$=\frac{-5}{2}$

So here,

First term (a) = 18

Last term $(l)=-49 \frac{1}{2}=\frac{-99}{2}$

Common difference $(d)=\frac{-5}{2}$

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

$a_{n}=a+(n-1) d$

So, for the last term,

$\frac{-99}{2}=18+(n-1) \frac{-5}{2}$

$\frac{-99}{2}=18+\left(\frac{-5}{2}\right) n+\frac{5}{2}$

$\frac{5}{2} n=18+\frac{5}{2}+\frac{99}{2}$

$\frac{5}{2} n=18+\frac{104}{2}$

$n=28$

Now, using the formula for the sum of n terms, we get

$S_{n}=\frac{28}{2}\left[2 \times 18+(28-1)\left(\frac{-5}{2}\right)\right]$

$S_{n}=14\left[36+27\left(\frac{-5}{2}\right)\right]$

 

$S_{n}=-441$

Therefore, the sum of the A.P is $S_{n}=-441$.