Question

If the nth term of an arithmatic progression is 3n+5, what is the sum of the 'n' terms?

Answer 1

Tn=3n+5 = 8+3n-3 = 8+(n-1)*3. therefore a=8 and d=3.
Hence Sn=n/2(2a+(n-1)d) = n/2(16+3n-3) = n/2(13+3n) = (13n+3n^2)/2.