Sequences and Series: Common Mistakes and Fixes (5)

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Question

The sum of the first nn terms of an AP is Sn=3n2+5nS_n = 3n^2 + 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)=8a_1 = S_1 = 3(1)^2 + 5(1) = 8.

S2=3(4)+5(2)=22S_2 = 3(4) + 5(2) = 22. So a1+a2=22a_1 + a_2 = 22, giving a2=228=14a_2 = 22 - 8 = 14.

d=a2a1=148=6d = a_2 - a_1 = 14 - 8 = 6.

For an AP, an=a1+(n1)da_n = a_1 + (n - 1)d. So:

a20=8+19×6=8+114=122a_{20} = 8 + 19 \times 6 = 8 + 114 = 122

S20=3(400)+5(20)=1300S_{20} = 3(400) + 5(20) = 1300. S19=3(361)+5(19)=1083+95=1178S_{19} = 3(361) + 5(19) = 1083 + 95 = 1178.

a20=13001178=122a_{20} = 1300 - 1178 = 122. ✓

a1=8a_1 = 8, d=6d = 6, a20=122a_{20} = 122.

Why This Works

The relation an=SnSn1a_n = S_n - S_{n-1} recovers the nn-th term from the cumulative sum. For the first term, we use a1=S1a_1 = S_1 directly. The common difference is then just a2a1a_2 - a_1.

If SnS_n is a quadratic in nn, the sequence is an AP — the coefficient of n2n^2 in SnS_n equals d/2d/2, and the linear coefficient relates to a1d/2a_1 - d/2. This shortcut works only for APs.

Alternative Method

Match coefficients directly. For an AP, Sn=(n/2)(2a+(n1)d)=(d/2)n2+(ad/2)nS_n = (n/2)(2a + (n-1)d) = (d/2)n^2 + (a - d/2)n. Comparing with 3n2+5n3n^2 + 5n: d/2=3d/2 = 3 so d=6d = 6; ad/2=5a - d/2 = 5 so a=8a = 8. Faster, no S2S_2 computation needed.

Common Mistake

Treating an=Sn/na_n = S_n / n. That would only give the average of the first nn terms, not the nn-th term. The correct relation is an=SnSn1a_n = S_n - S_{n-1} for n2n \geq 2.

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