Question

Find the indicated term in each arithmetic sequence.

$70\text{th term of}2\sqrt{5},4\sqrt{5},6\sqrt{5},\dots$

Short answer

The 70th term of the sequence is$140\sqrt{5}.$

Step-by-step solution

Step 1. Given information

$\text{Thegivensequenceis}2\sqrt{5},4\sqrt{5},6\sqrt{5},\dots$

We have to find the 70th term of the sequence.

Step 2. Calculating the common difference.

$\begin{array}{l}\Rightarrow a_1=2\sqrt{5},a_2=4\sqrt{5},a_3=6\sqrt{5}\\ \Rightarrow a_2-a_1=4\sqrt{5}-2\sqrt{5}\\ =2\sqrt{5}\\ \Rightarrow a_3-a_2=6\sqrt{5}-4\sqrt{5}\\ =2\sqrt{5}\\ \Rightarrow \text{commondifference,}d=2\sqrt{5}\end{array}$

Step 3. Calculating the required term.

$\begin{array}{l}\text{The}{n}^{th}\text{termofasequenceisgivenas}\\ a_n=a_1+\left(n-1\right)d\\ \text{Here,}\\ a\text{isthefirsttermofthesequence.}\\ d\text{isthecommondifference.}\\ \text{Thus,}\\ \Rightarrow a_{70}=2\sqrt{5}+\left(70-1\right)\left(2\sqrt{5}\right)\\ =2\sqrt{5}+69\cdot \left(2\sqrt{5}\right)\\ =2\sqrt{5}+138\sqrt{5}\\ =140\sqrt{5}\end{array}$