Problem: For every n the sum of n terms of an arithmetic progression is 2n+3n2. The rth term is:
Answer Choices:
A. 3r2
B. 3r2+2r
C. 6r−1
D. 5r+5
E. 6r+2
Solution:
The rth term equals Sr−Sr−1 and Sr=2r+3r2 and Sr−1=2(r−1)+3(r−1)2=3r2−4r+1 ∴ the rth term equals 6r−1
Let the rth term be ur; then Sr=2r(a+ur)=2r+3r2∴a+ur=4+6r
But a=S1=2⋅1+3⋅12=5∴ur=4+6r−5=6r−1
or
a=S1=5,S2=16∴u2=S2−S1=11, but u2=a+d∴d=6
∴ur=a+(r−1)d=5+(r−1)(6)=6r−1