Question
The sum of the first n terms of an AP is Sn=3n2+5n. Find (a) the first term, (b) the common difference, and (c) the 20th term.
Solution — Step by Step
a1=S1=3(1)2+5(1)=8.
S2=3(4)+5(2)=22. So a1+a2=22, giving a2=22−8=14.
d=a2−a1=14−8=6.
For an AP, an=a1+(n−1)d. So:
a20=8+19×6=8+114=122
S20=3(400)+5(20)=1300. S19=3(361)+5(19)=1083+95=1178.
a20=1300−1178=122. ✓
a1=8, d=6, a20=122.
Why This Works
The relation an=Sn−Sn−1 recovers the n-th term from the cumulative sum. For the first term, we use a1=S1 directly. The common difference is then just a2−a1.
If Sn is a quadratic in n, the sequence is an AP — the coefficient of n2 in Sn equals d/2, and the linear coefficient relates to a1−d/2. This shortcut works only for APs.
Alternative Method
Match coefficients directly. For an AP, Sn=(n/2)(2a+(n−1)d)=(d/2)n2+(a−d/2)n. Comparing with 3n2+5n: d/2=3 so d=6; a−d/2=5 so a=8. Faster, no S2 computation needed.
Common Mistake
Treating an=Sn/n. That would only give the average of the first n terms, not the n-th term. The correct relation is an=Sn−Sn−1 for n≥2.