Find the sum of first 15 terms of an A.P. whose nth term is 9-5n.
PLEASE ANSWER STEPWISE :)
Answers
Answered by
2
I hope it will help you.
Attachments:
Answered by
16
n= 9-5n
taking n= 1 , 2, ,3...,
case (i) n=1
=9-5(1) =9-5 = 4
taking n=2
=9-5(2) = 9-10 = (-1)
taking n = 3
=9-5(3) =9-15 = -6
here A1=4 , A2=-1 , A3= -6
so, A=4 , d =-5 n=15
Sn=n/2 [2a+(n-1)d]
S15=15/2[2(4) +( 15 - 1 ) -5]
=15/2[8 + [14]-5]
=15/2[8+-70]
=15/2[(-62]
=15[-31]
=-465
plszzz mark as brainliest answer !!!
taking n= 1 , 2, ,3...,
case (i) n=1
=9-5(1) =9-5 = 4
taking n=2
=9-5(2) = 9-10 = (-1)
taking n = 3
=9-5(3) =9-15 = -6
here A1=4 , A2=-1 , A3= -6
so, A=4 , d =-5 n=15
Sn=n/2 [2a+(n-1)d]
S15=15/2[2(4) +( 15 - 1 ) -5]
=15/2[8 + [14]-5]
=15/2[8+-70]
=15/2[(-62]
=15[-31]
=-465
plszzz mark as brainliest answer !!!
Similar questions