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

Given that;

The sixth and twelfth are as follows

`\begin{gathered} a_6=(3)/(2) \\ a_(12)=(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 \\ (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 \\ (5)/(2)=a_1+11d...(2) \end{gathered}`

Solving the equations simultaneously

`\begin{gathered} (3)/(2)=a_1+5d \\ a_1=(3)/(2)-5d...(3) \\ Substitute\text{ for a}_1\text{ into equation \lparen2\rparen} \\ (5)/(2)=a_1+11d \\ (5)/(2)=((3)/(2)-5d)+11d \\ (5)/(2)=(3)/(2)-5d+11d \\ (5)/(2)=(3)/(2)+6d \\ Collect\text{ like terms} \\ 6d=(5)/(2)-(3)/(2) \\ 6d=(5-3)/(2)=(2)/(2)=2 \\ 6d=1 \\ Divide\text{ both sides by 6} \\ (6d)/(6)=(1)/(6) \\ d=(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=(3)/(2)-5d \\ a_1=(3)/(2)-5((1)/(6))=(3)/(2)-(5)/(6) \\ a_1=(9-5)/(6)=(4)/(6)=(2)/(3) \\ a_1=(2)/(3) \end{gathered}`

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

`\begin{gathered} a_1=(2)/(3) \\ d=(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=(2)/(3)+(n-1)(1)/(6) \end{gathered}`

Hence, the explicit formula is

`a_(n)=(2)/(3)+(n-1)(1)/(6)`