1) The first term of arithmetic progression is -20 and the sum of it's term is 250. find it's last term if the number of it's terms is 10
2) Find the fifth term from the arithmetic progression -12, -9, -6 hence find the sum of it's first fifty terms.
3) The sum of the first terms in Arithmetic progression is equal to -24, and the sum of the last six terms equal to 30. if the number of it's term is 12. find a) it's common different
b) The first term
c) the last term
6 answers
Answer
3)a=-12
d=3
t(5)=a+(n-1)d
=-12+(5-1)3
=0
s(50)=3625
hope it would be a help. :)
d=3
t(5)=a+(n-1)d
=-12+(5-1)3
=0
s(50)=3625
hope it would be a help. :)
U2= a=8/9 U2=8/9 Un= arn-1 if n=2 U2= arn7-1 8/9=2(r6)
1) a=-20
Sn=250
. n=10
Sn=n/2(2a+(n-1)d)
250=10/2(2(-20)+(10-1)d)
250=5((-40)+(9)d)
250=-200+45d
250+200=45d
d=10
Un=a+(n-1)d
Un=-20*(10-1)10
Un=-20+900
Un=880
::the last term=880
Sn=250
. n=10
Sn=n/2(2a+(n-1)d)
250=10/2(2(-20)+(10-1)d)
250=5((-40)+(9)d)
250=-200+45d
250+200=45d
d=10
Un=a+(n-1)d
Un=-20*(10-1)10
Un=-20+900
Un=880
::the last term=880
2) a=-12
. Un=-6
d=-3
n=5
S5=5/2(2(-12)+(5-1)-3)
2.5(-24)+-12)
. 2.5(-36)
S5=-90
. Un=-6
d=-3
n=5
S5=5/2(2(-12)+(5-1)-3)
2.5(-24)+-12)
. 2.5(-36)
S5=-90
1) Sn= n/2 (a + l )
250=10/2 (-20+l)
250=(-200+10l)/2
250 x 2= -200+10L
500=-200+10L
500+200= 10L
700=10L
L= 700/10
L=70
250=10/2 (-20+l)
250=(-200+10l)/2
250 x 2= -200+10L
500=-200+10L
500+200= 10L
700=10L
L= 700/10
L=70