Exercise 201Calculus › Single-variable calculus › SequencesHigh schoolAdded 2 Nov 2026

Arithmetic and geometric sequences

#arithmetic sequence#geometric sequence#sums

Statement

Let $(u_n)$ be arithmetic with $u_0=2$ and common difference $4$ . Compute $u_{10}$ and $S_n=\sum_{k=0}^nu_k$ .
Let $(v_n)$ be geometric with $v_0=3$ and common ratio $\frac12$ . Compute $v_5$ , $T_n=\sum_{k=0}^nv_k$ , and $\lim_{n\to\infty}T_n$ .

Hint

Arithmetic: $u_n=u_0+nr$ and $\sum_{k=0}^nu_k=(n+1)\frac{u_0+u_n}2$ . Geometric: $v_n=v_0q^n$ and $\sum_{k=0}^nv_k=v_0\frac{1-q^{n+1}}{1-q}$ for $q\ne1$ .

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Solution

1. $u_n=2+4n$ , so $u_{10}=42$ and
$S_n=(n+1)\frac{2+(2+4n)}2=(n+1)(2+2n)=2(n+1)^2.$
(Check: $S_1=2+6=8=2\cdot4$ .)

2. $v_n=3\left(\frac12\right)^n$ , so $v_5=\frac3{32}$ and
$T_n=3\cdot\frac{1-2^{-(n+1)}}{1-\frac12}=6\left(1-2^{-(n+1)}\right).$
Since $2^{-(n+1)}\to0$ , $T_n\to6$ .

Remark. Adding infinitely many positive numbers can give a finite sum: this is the first example of a convergent series.

Arithmetic and geometric sequences

Statement

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