helppp

The mean and range of ten consecutive terms in an arithmetic progression are all 82.
What is the first term in the sequence?

\(\text{Let $a$ is the first term} \\ \text{Let $b$ is the last term}\)

\(\begin{array}{|lrcll|} \hline \text{mean:} & \dfrac{a+b}{2} &=& 82 \quad & | \quad \cdot 2 \\\\ & a+b &=& 2\cdot 82 \qquad (1) \\ \\ \text{range:} & b-a &=& 82 \qquad (2) \\ \hline \\ (1)-(2):& (a+b) - (b-a) &=& 2\cdot 82 - 82 \\ & a+b - b+a &=& 82 \\ & a+a &=& 82 \\ & 2a &=& 82 \quad & | \quad : 2 \\ & \mathbf{ a } & \mathbf{=} & \mathbf{41} \\ \hline \end{array}\)

The first term in the sequence 41

If the FIRST term is 41 and the LAST term is 123, how can the "arithmetic progression" consist of " ten consecutive terms"??? In fact the "common difference" would have to be 9 1/9 !!!!!!.

If you mean that the "range and mean combined =82", then the AP could be:

68.50, 69.50, 70.50, 71.50, 72.50, 73.50, 74.50, 75.50, 76.50, 77.50 =730/10 =73 - the mean

77.5 - 68.50 = 9 - the range

73 + 9 = 82 combined range and mean!!!!