SOLUTION: I need detail solution of this question: The first term of an arithmetic progression is 8. The sum of the first 10 terms of this progression divided by the sum of the first 5 ter

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Question 823755: I need detail solution of this question:
The first term of an arithmetic progression is 8. The sum of the first 10 terms of this progression
divided by the sum of the first 5 terms is 3.25. How many terms of this progression add up
to 1620?
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
I need detail solution of this question:
The first term of an arithmetic progression is 8. 
a1=8
The sum of the first 10 terms of this progression...
Sn = n%2F2[2a1 + (n-1)d]

S10 = 10%2F2[2(8) + (10-1)d] =

                 5[16 + 9d] =

                 80 + 45d
...divided by...
...the sum of the first 5 terms...
Sn = n%2F2[2a1 + (n-1)d]

S5 = 5%2F2[2(8) + (5-1)d] =

                5%2F2[16 + 4d] =

                 40 + 10d
...is 3.25:

%2880+%2B+45d%29%2F%2840+%2B+10d%29 = 3.25  [Note: 3.25=3%261%2F4=13%2F4 ]

%2880+%2B+45d%29%2F%2840+%2B+10d%29 = 13%2F4

Cross-multiply:

4(80 + 45d) = 13(40 + 10d)
 320 + 180d = 520 + 130d
        50d = 200
          d = 4
How many terms of this progression add up to 1620?
Sn = n%2F2[2a1 + (n-1)d]

1620 = n%2F2[2(8) + (n-1)4]

1620 = n%2F2[16 + 4(n-1)]

11620 = n%2F2[16 + 4n - 4]

1620 = n%2F2[12 + 4n]

1620 = 6n + 2nē

   0 = 2nē + 6n - 1620

Divide every term by 2

   0 = nē + 3n - 810

   0 = (n-27)(n+30)

    n-27=0;    n+30=0
       n=27;      n=-30 (ignore)

Answer:  27 terms add up to 1620.

Edwin