Avg of First n Terms ⇒ Sₙ and tₙ — One-Page Cheat Code
October 31, 2025 2025-10-31 14:53Avg of First n Terms ⇒ Sₙ and tₙ — One-Page Cheat Code
Average of First \(n\) Terms ⇒ \(S_n\) and \(t_n\) — One-Page Cheat Code
Core idea: If the average of the first \(n\) terms is \(A_n\), then the sum \(S_n=nA_n\). The \(n^\text{th}\) term is the difference of consecutive sums: \(t_n=S_n-S_{n-1}\). When \(S_n\) is quadratic, the sequence is an AP. When \(t_n\) is linear, \(S_n\) is quadratic (with \(\tfrac12\) on the \(n^2\) coefficient).
1) From Average to Sum and Term
\[
A_n=\text{Average of first }n\text{ terms} \quad\Longrightarrow\quad
S_n=nA_n,\qquad t_n=S_n-S_{n-1}.
\]
Why it works: By definition \(A_n=\dfrac{t_1+t_2+\cdots+t_n}{n}=\dfrac{S_n}{n}\).
The \(n^\text{th}\) term always equals the increment in total:
\(\;t_n=S_n-S_{n-1}\) with the convention \(S_0=0\).
2) Two Must-Know Cases
(A) Linear term: \(t_n = an + b\)
\[
S_n=\sum_{k=1}^{n}(ak+b)
= a\frac{n(n+1)}{2} + bn
= \frac{a}{2}n^2+\Big(\frac{a}{2}+b\Big)n.
\]
\[
A_n=\frac{S_n}{n}=\frac{a}{2}(n+1)+b.
\]
This is an AP with first term \(t_1=a+b\) and common difference \(d=a\).
(B) Quadratic sum: \(S_n=an^2+bn\)
\[
t_n=S_n-S_{n-1}=a(2n-1)+b = (2a)n + (b-a).
\]
Hence \(t_n\) is an AP with first term \(t_1=a+b\) and common difference \(d=2a\). Also \(A_n=\dfrac{S_n}{n}=an+b\) (linear in \(n\)).
AP mean shortcut: For any AP, the average of first \(n\) terms equals the average of first and \(n^\text{th}\) terms:
\[
A_n=\frac{t_1+t_n}{2}.
\]
3) Worked Examples
Example 1 — \(A_n\) given (linear)
Given \(A_n=3n+2\). Find a formula for \(t_n\) and compute \(t_{10}\).
Solution
\[ S_n=nA_n=n(3n+2)=3n^2+2n,\quad S_{n-1}=3(n-1)^2+2(n-1)=3n^2-4n+1. \] \[ t_n=S_n-S_{n-1}=(3n^2+2n)-(3n^2-4n+1)=6n-1 \;\Rightarrow\; t_{10}=59. \]Example 2 — \(t_n\) given (linear)
Given \(t_n=5n-1\). Find \(S_n\) and \(A_n\).
Solution
\[ S_n=\sum_{k=1}^n(5k-1)=5\frac{n(n+1)}{2}-n=\frac{5}{2}n^2+\frac{3}{2}n, \quad A_n=\frac{S_n}{n}=\frac{5}{2}(n+1)-1. \]Example 3 — \(S_n\) quadratic
Given \(S_n=4n^2-3n\). Find \(t_1\), common difference \(d\), and \(t_n\).
Solution
From \(S_n=an^2+bn\) with \(a=4,b=-3\): \(t_1=a+b=1\), \(d=2a=8\), and \[ t_n=S_n-S_{n-1}=4(2n-1)-3=8n-7. \]Example 4 — Non-constant start, then stable
Let \(A_n=7-\dfrac{12}{n}\). Find \(t_1\) and \(t_n\) for \(n\ge2\).
Solution
\(S_n=nA_n=7n-12\). \(t_1=S_1-S_0=(7-12)-0=-5\). For \(n\ge2\): \(t_n=S_n-S_{n-1}=(7n-12)-[7(n-1)-12]=7\). So the sequence is \(-5,7,7,7,\dots\)4) Quick Review (What to Remember)
- \(A_n\) known ⇒ \(S_n=nA_n\) and \(t_n=S_n-S_{n-1}\).
- If \(t_n=an+b\), then \(S_n=\dfrac{a}{2}n(n+1)+bn\) and it’s an AP with \(t_1=a+b,\; d=a\).
- If \(S_n=an^2+bn\), then \(t_n=(2a)n+(b-a)\) and it’s an AP with \(t_1=a+b,\; d=2a\).
- AP average: \(A_n=\dfrac{t_1+t_n}{2}\).
- Always set \(S_0=0\) for clean edge cases.
5) Practice for Revision
Set A — Direct Use of \(S_n=nA_n\) and \(t_n=S_n-S_{n-1}\)
-
If \(A_n=2n+3\), find \(t_n\) and \(t_{12}\).
Answer
\(S_n=2n^2+3n\Rightarrow t_n=4n-1,\; t_{12}=47\). -
If \(A_n=5-\dfrac{9}{n}\), find \(t_1\) and \(t_8\).
Answer
\(S_n=5n-9\Rightarrow t_1=-4\). For \(n\ge2\), \(t_n=5\) ⇒ \(t_8=5\). -
A sequence has \(A_n=n^2\). Find \(t_n\).
Answer
\(S_n=n\cdot n^2=n^3\Rightarrow t_n=n^3-(n-1)^3=3n^2-3n+1\). -
If \(A_n=\dfrac{n+1}{2}\), identify the sequence.
Answer
\(S_n=\dfrac{n(n+1)}{2}\Rightarrow t_n=n\). (Natural numbers)
Set B — Linear \(t_n\)
-
\(t_n=7n+2\). Find \(S_n\) and \(A_n\). What are \(t_1\) and \(d\)?
Answer
\(S_n=7\frac{n(n+1)}{2}+2n=\frac{7}{2}n^2+\frac{11}{2}n\), \(A_n=\frac{7}{2}(n+1)+2\), \(t_1=9\), \(d=7\). -
\(t_n=12-3n\). Find the smallest \(n\) for which \(t_n\le 0\).
Answer
\(12-3n\le 0 \Rightarrow n\ge 4\). First non-positive term at \(n=4\).
Set C — Quadratic \(S_n\)
-
\(S_n=3n^2+5n\). Find \(t_1\), \(d\), and \(t_{20}\).
Answer
\(t_1=3+5=8\), \(d=2\cdot3=6\), \(t_n=6n+2\Rightarrow t_{20}=122\). -
\(S_n=an^2+bn\) yields \(t_8=50\) and \(t_{13}=80\). Find \(a,b\).
Answer
\(t_n=(2a)n+(b-a)\). \(16a+(b-a)=50\Rightarrow 15a+b=50\). \(26a+(b-a)=80\Rightarrow 25a+b=80\). Subtract: \(10a=30\Rightarrow a=3\). Then \(b=5\).
Set D — Mixed
-
If \(A_n=4n+c\) and \(t_{10}=79\), find \(c\).
Answer
\(S_n=4n^2+cn\Rightarrow t_n=8n+(c-4)\). \(t_{10}=80+c-4=79\Rightarrow c=3\). -
For \(t_n=an+b\), the average of first \(n\) terms equals \(11\) when \(n=5\) and equals \(15\) when \(n=9\). Find \(a,b\).
Answer
\(A_n=\frac{a}{2}(n+1)+b\). \( \frac{a}{2}\cdot6+b=11 \Rightarrow 3a+b=11\). \( \frac{a}{2}\cdot10+b=15 \Rightarrow 5a+b=15\). Subtract: \(2a=4\Rightarrow a=2\). Then \(b=5\). -
A sequence satisfies \(A_n=6n-1\). Show it is an AP and find \(d\).
Answer
\(S_n=6n^2-n\Rightarrow t_n=S_n-S_{n-1}=12n-7\). Linear \(t_n\) ⇒ AP with constant difference \(d=12\). -
Prove: If \(A_n\) is linear in \(n\), then \((t_n)\) is an AP.
Proof sketch
Linear \(A_n=pn+q\Rightarrow S_n=nA_n=pn^2+qn\) (quadratic). Then \(t_n=S_n-S_{n-1}=2pn+(q-p)\) which is linear in \(n\) ⇒ AP.
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