Question

What is the explicit rule for the arithmeticsequence?19/1/2ar- 1 + (x-1)/² * a -1 + (x - 1²/an =3anArn=1+(m-12+ (n − 1) /2/2 + (12-1)/1/1/ORETRY✔done

Answer 1

Solution:

Given that;

The sixth and twelfth are as follows

$\begin{gathered} a_6=\frac{3}{2} \\ a_{12}=\frac{5}{2} \end{gathered}$

Applying the Arithmetic progression formula below

$a_n=a_1+(n-1)d$

For the sixth term

$\begin{gathered} a_6=a_1+(6-1)d \\ a_6=a_1+5d \\ \frac{3}{2}=a_1+5d...(1) \end{gathered}$

For the twelfth term

$\begin{gathered} a_{12}=a_1+(12-1)d \\ a_{12}=a_1+11d \\ \frac{5}{2}=a_1+11d...(2) \end{gathered}$

Solving the equations simultaneously

$\begin{gathered} \frac{3}{2}=a_1+5d \\ a_1=\frac{3}{2}-5d...(3) \\ Substitute\text{ for a}_1\text{ into equation \lparen2\rparen} \\ \frac{5}{2}=a_1+11d \\ \frac{5}{2}=(\frac{3}{2}-5d)+11d \\ \frac{5}{2}=\frac{3}{2}-5d+11d \\ \frac{5}{2}=\frac{3}{2}+6d \\ Collect\text{ like terms} \\ 6d=\frac{5}{2}-\frac{3}{2} \\ 6d=\frac{5-3}{2}=\frac{2}{2}=2 \\ 6d=1 \\ Divide\text{ both sides by 6} \\ \frac{6d}{6}=\frac{1}{6} \\ d=\frac{1}{6} \end{gathered}$

Where, the common difference is 1/6,

Substitute into equation (3) to find the first term, a₁

$\begin{gathered} a_1=\frac{3}{2}-5d \\ a_1=\frac{3}{2}-5(\frac{1}{6})=\frac{3}{2}-\frac{5}{6} \\ a_1=\frac{9-5}{6}=\frac{4}{6}=\frac{2}{3} \\ a_1=\frac{2}{3} \end{gathered}$

Hence, the first term, a₁ and common difference, d, are

$\begin{gathered} a_1=\frac{2}{3} \\ d=\frac{1}{6} \end{gathered}$

Where the first term, a₁, is 2/3 and the common difference, d, is 1/6,

The explicit formula is

$\begin{gathered} a_n=a_1+(n-1)d \\ a_n=\frac{2}{3}+(n-1)\frac{1}{6} \end{gathered}$

Hence, the explicit formula is

$a_{n}=\frac{2}{3}+(n-1)\frac{1}{6}$