The sum of the first nine terms of ap is 72 and the sum of the next four term is 71.find the ao
Answers
Step-by-step explanation:
Term An of the arithmetic series An = A0 +(n-1)d
where A0 is the 1st term and d is the difference.
We know the sum of the 1st 9 terms is 72
That would be 9*A0 + (d+2d+3d . . .8d) = 10*A0 + 36*d
So, 72 = 9*A0 + 36*d
The next 4 terms (10 through 13) = 71,
so 4*A0 + (9d + 10d + 11d+ 12d) = 71, that is
4*A0 + 42*d = 71
Now we have two simultaneous equations to solve:
9*A0 + 36*d = 72
4*A0 + 42*d = 71
Multiply the 1st by 4 and the 2nd by 9, then Subtract the 1st from the second
-36*A0 -144*d = -288
36*A0 + 378*d = 639 234*d = 351, so d = 3/2
Plugging back into 4*A0 + 42*d = 71
4*A0 +42*(3/2) = 71 solves as A0 = 2
The AP is An = 2+(n-1)*(3/2)
To check this, use a TI-84 and input the following:
Sum(seq(2+(3/2)*(x-1),x,1,9)) And it will give you 72
Then the next 4
Sum(seq(2+(3/2)*(x-1),x,10,13)) And it will give you 71
The AP is An = 2+(n-1)*(3/2)