+0  
 
0
37
2
avatar

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.

 Jul 8, 2020
 #1
avatar
0

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

 Jul 8, 2020
 #2
avatar
0

Sorry young person! I only summed up the first 6 terms. Here are all the 10 terms:

 

S = 10/2 * [2*(293/3) + (28/3) * 9], solve for S

 

S =4190 / 3 =1396 + 2/3

Guest Jul 8, 2020

26 Online Users

avatar