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