Algebraic Patterns Practise #2

\(\begin{array} {} & a_1 = 2 & = 2 \\ & a_2 = 2 + 7 & = 9 \\ & a_3 = 2+ 2 \times 7 & = 16 \\ & a_4 = 2+ 3 \times 7 & = 23 \\ & \cdots \ = \quad \cdots & = \cdots \\ \implies & a_n = 2+ 7(n-1) & = 7n-5 \\ & \cdots \ = \quad \cdots & = \cdots \\ \implies & a_{100} = 2+ 7(100-1) & = \boxed{695} \end{array} \)

If we are bored with calculations we can just notice that each number in this sequence is of the form $7n+2$ and since only $695$ is of this form it must be the answer.

Start: Just forget 2 in each term (i.e. subtract 2). You will immediately observe it to be a table of 7 with terms 0, 7, 14..., 7*(n-1) for nth term.

Answer: 100th term will be 7*(100 -1) + 2 (remember we subtracted, in the end we can't forget to add it).

Simple, :)

This is an algebraic pattern, so let's find an appropriate algebraic expression to be able to solve for the 100th term.

Let n be the term number.

This algebraic pattern increases by 7 from term to term, so we will be able to get 7n as our base. However, in order to get to the term value, we need to subtract 5. Now we have 7n-5. To solve this question, we simply have to substitute n with 100 and solve.

7n-5 = 7(100)-5 = 700-5 = 695

Therefore, the 100th term in this sequence is 695.