Find the sum:
(i) $2+4+6 \ldots+200$
(ii) $3+11+19+\ldots+803$
(iii) $(-5)+(-8)+(-11)+\ldots+(-230)$
(iv) $1+3+5+7+\ldots+199$
(v) $7+10 \frac{1}{2}+14+\ldots+84$
(vi) $34+32+30+\ldots+10$
(vii) $25+28+31+\ldots+100$
(viii) $18+15 \frac{1}{2}+13+\ldots+\left(-49 \frac{1}{2}\right)$
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 $S_{n}=-441$.
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