Thinking Mathematically (6th Edition)

The provided sequence is:$-10,-6,-2,2\ldots$ Here, ${{a}_{1}}=-10,{{a}_{2}}=-6,{{a}_{3}}=-2\ \text{and}\ {{a}_{4}}=2$ As, \begin{align} & {{a}_{2}}-{{a}_{1}}=-6+10 \\ & =4 \\ & {{a}_{3}}-{{a}_{2}}=-2+6 \\ & =4 \\ & {{a}_{4}}-{{a}_{3}}=2+2 \\ & =4 \end{align} This implies common difference (d) between two consecutive terms is: 4 Therefore, The provided sequence is arithmetic. Here ${{a}_{1}}$ and ${{a}_{n}}$is first and the nth term of the provided A.P and n is natural number. To find the sum of the first ten terms, 10th and 1st term of the sequence are needed. To find 10th term of the provided sequence formula to be used is:${{a}_{n}}={{a}_{1}}+\left( n-1 \right)d$ Put $n=10,{{a}_{1}}=-2,d=4$ in the above formula, the 10th term of the sequence is: \begin{align} & {{a}_{10}}=-10+\left( 10-1 \right)4 \\ & =-10+36 \\ & =26 \end{align} Therefore,the required sum of the provided sequence is: \begin{align} & {{s}_{10}}=\frac{10}{2}\left( {{a}_{1}}+{{a}_{10}} \right) \\ & =\frac{10}{2}\left( -10+26 \right) \\ & =5\left( 16 \right) \\ & =80 \end{align} The provided sequence is Arithmetic and sum of first 10 terms is 80.