Question

1. in an arithmetic sequence, ( u_2 = 5 ) and ( u_3 = 11 ).

a) find the common difference.
b) find the first term.
c) find the sum of the first 20 terms.

Explanation:

Part (a)

Step1: Recall arithmetic sequence formula

In an arithmetic sequence, the \(n\)-th term is given by \(u_n = u_1+(n - 1)d\), where \(d\) is the common difference. Also, \(u_{n+1}=u_n + d\). So, to find the common difference \(d\), we can use the formula \(d=u_{n + 1}-u_n\).

Step2: Calculate the common difference

Given \(u_2 = 5\) and \(u_3=11\), using \(d = u_3 - u_2\), we substitute the values: \(d=11 - 5\)
\(d = 6\)

Step1: Recall the arithmetic sequence formula

The formula for the \(n\)-th term of an arithmetic sequence is \(u_n=u_1+(n - 1)d\). We know \(u_2 = 5\), \(n = 2\) and \(d = 6\) (from part (a)).

Step2: Solve for \(u_1\)

Substitute \(n = 2\), \(u_2=5\) and \(d = 6\) into \(u_n=u_1+(n - 1)d\):
\(5=u_1+(2 - 1)\times6\)
\(5=u_1 + 6\)
Subtract \(6\) from both sides: \(u_1=5 - 6=- 1\)

Step1: Recall the sum formula for arithmetic sequence

The sum of the first \(n\) terms of an arithmetic sequence is given by \(S_n=\frac{n}{2}(2u_1+(n - 1)d)\) or \(S_n=\frac{n}{2}(u_1 + u_n)\). We know \(n = 20\), \(u_1=-1\) and \(d = 6\). First, we can find \(u_{20}\) using \(u_n=u_1+(n - 1)d\) or use the first formula.

Step2: Calculate \(u_{20}\) (optional, but can be used)

Using \(u_n=u_1+(n - 1)d\), with \(n = 20\), \(u_1=-1\) and \(d = 6\):
\(u_{20}=-1+(20 - 1)\times6=-1 + 19\times6=-1+114 = 113\)

Step3: Calculate the sum \(S_{20}\)

Using \(S_n=\frac{n}{2}(u_1 + u_n)\), with \(n = 20\), \(u_1=-1\) and \(u_{20}=113\):
\(S_{20}=\frac{20}{2}(-1 + 113)\)
\(S_{20}=10\times112\)
\(S_{20}=1120\)

Alternatively, using \(S_n=\frac{n}{2}(2u_1+(n - 1)d)\):
\(S_{20}=\frac{20}{2}(2\times(-1)+(20 - 1)\times6)\)
\(S_{20}=10(-2 + 114)\)
\(S_{20}=10\times112 = 1120\)

Answer:

The common difference is \(6\).

Part (b)