There are ten numbers in a certain arithmetic progression. The sum of first three terms is 321. The sum of last three numbers is 405. Find the sum of all the ten numbers.
3/2 * [2*F + 2D]=321, [3F + 12D] =405, solve for D, F F =293/3 - This is your first term D =28/3 - This is the common difference. Sum = 6/2 * [2*(293/3) + (28/3) * 5] Sum = 726
