Originally Posted by
HallsofIvy Not a whole lot! My question now is why you have $\displaystyle s_2$ and $\displaystyle a_8$? You are giving us one term from each of two sequences!
Or did your finger just slip on the key board and it is $\displaystyle s_2$ and $\displaystyle s_8$ or $\displaystyle a_2$ and $\displaystyle a_8$?
Knowing that $\displaystyle a_2= 356$ and $\displaystyle a_8= 83$ tells us that a+ 2d= 356 and a+ 8d= 83. Subtracting the first equation from the second, 6d= 273 so d= 45.5. Now, a+2d= a+ 91= 356 so a= 356- 91= 265. You could, now calculate the first ten terms and add.
Are you saying that I now must do the math and calculate the first 10 terms and then add? Can you please show me how this is done?
But arthmetic sequences have the very nice property that the "average" of an n consecutive terms is the same as the average of the first and last terms in the sequence. Here, the first term is 265 and the tenth term is [tex]a_{10}= 265- 10(45.5)= 265- 455= -190. The average of those two numbers is (265-190)/2= 75/2= 37.5 and that is the average of the first 10 numbers.