Duplicate Question
The question on this page has been marked as a duplicate question.
Original Question
find the sum of the first 12th terms of the sequence 8, 5, 2 ......Asked by Blessing
Find the sum of the first 12th terms of the sequence 2,5,8,11
Answers
Answered by
Anonymous
term n = a + d(n-1)
a = 2
d = 3
sum from 0 to n = (n/2) [ 2 a +(n-1)d ]
n = 12
sum = 6 [ 4 + 11*3] = 6 * 37
a = 2
d = 3
sum from 0 to n = (n/2) [ 2 a +(n-1)d ]
n = 12
sum = 6 [ 4 + 11*3] = 6 * 37
Answered by
Bolu
Is far final answer
Answered by
Suliat
222
Answered by
Authority
T12 =s=n/2 [2a+(n-1)d
a=2
d=5-2=3
S=12/2[2*2+(12-1)3]
[4+11*3]
[6*37]
= 222
a=2
d=5-2=3
S=12/2[2*2+(12-1)3]
[4+11*3]
[6*37]
= 222
Answered by
Rokeebat
227
Answered by
Viannie
227
Answered by
Precious
an=a1+(n-1)
a1=2
d=3
an=2+(n-1)×3
=2+3n-3
=3n-3
S=(n/2) (a1+an)
S=(12/2) [2+(3×12-1)]
6(2+35)
6=37
=222
a1=2
d=3
an=2+(n-1)×3
=2+3n-3
=3n-3
S=(n/2) (a1+an)
S=(12/2) [2+(3×12-1)]
6(2+35)
6=37
=222
Answered by
Bot
I apologize, but I couldn't quite understand the context of this response. Could you please provide more information or clarify your question?