Question 823755
I need detail solution of this question:
The first term of an arithmetic progression is 8. 
<pre>
a<sub>1</sub>=8
</pre>The sum of the first 10 terms of this progression...<pre>
S<sub>n</sub> = {{{n/2}}}[2a<sub>1</sub> + (n-1)d]

S<sub>10</sub> = {{{10/2}}}[2(8) + (10-1)d] =

                 5[16 + 9d] =

                 80 + 45d
</pre>...divided by...

...the sum of the first 5 terms...<pre>
S<sub>n</sub> = {{{n/2}}}[2a<sub>1</sub> + (n-1)d]

S<sub>5</sub> = {{{5/2}}}[2(8) + (5-1)d] =

                {{{5/2}}}[16 + 4d] =

                 40 + 10d
</pre>...is 3.25:<pre>

{{{(80 + 45d)/(40 + 10d)}}} = 3.25  [Note: 3.25={{{3&1/4}}}={{{13/4}}} ]

{{{(80 + 45d)/(40 + 10d)}}} = {{{13/4}}}

Cross-multiply:

4(80 + 45d) = 13(40 + 10d)
 320 + 180d = 520 + 130d
        50d = 200
          d = 4 
</pre>How many terms of this progression add up to 1620?<pre>
S<sub>n</sub> = {{{n/2}}}[2a<sub>1</sub> + (n-1)d]

1620 = {{{n/2}}}[2(8) + (n-1)4]

1620 = {{{n/2}}}[16 + 4(n-1)]

11620 = {{{n/2}}}[16 + 4n - 4]

1620 = {{{n/2}}}[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</pre>