T12 = T5 + 7d, so d = (103-82)/7 = 3
Now you know a = T5-4d = 82-4*3 = 70
So, S21 = 21/2 (2*70 + 20*3) = 2100
The fifth term of an AP is 82 and the 12th term is 103.find the first term and the common difference.
The sum of the first 21 terms
2 answers
S21=21/2(2(70)+(21-1)3
10.5(2(70)+(20)3)
10.5(140+60)
10.5(200)
=2100
10.5(2(70)+(20)3)
10.5(140+60)
10.5(200)
=2100