Question 1088240
<pre>
The multiples of 8 form an arithmetic sequence with common
difference 8.  So d = 8

100/8 = 12.5
The next whole number is 13, so the 13th multiple of 8, or 
13×8 = 104 is the smallest whole number divisible by 8 which 
is greater than 100. 

500/8 = 62.6
The previous whole number is 62, so the 62nd multiple of 8, or
62×8 = 496 is the largest whole number divisible by 8 which is 
less than 500.

From the 13th to the 62nd multiple of 8 to the 62nd multiple
of 8 is how many multiples of 8?

We can tell by subtracting 12 from both 13 and 60 to make the 
13th multiple of 8 the 1st one we are to consider. 12 subtracted 
from 62 is 50.  So there are 50 multiples of 8 in our sequence. 
So n=50

You can also find n this way:

a<sub>1</sub> = 104 and a<sub>n</sub> = 496

{{{a[n]=a[1]+(n-1)d}}}
{{{496=104+(n-1)8}}}
{{{392=8(n-1)}}}
{{{392=8n-8}}}
{{{400=8n}}}
{{{50=n}}}

The formula for the sum is

{{{S[n]=expr(n/2)(a[1]+a[n])}}}

{{{S[50]=expr(50/2)(104+496)}}}

{{{S[50]=25(600)}}}

{{{S[50]=15000}}}

------------------

{{{matrix(1,12,

common,
ratio,
""="",
r,
""=""
,matrix(1,2,2nd,term)/matrix(1,2,1st,term),
""="",
matrix(1,2,3rd,term)/matrix(1,2,2nd,term),
""="",
x/144,
""="",
64/x
)}}}

We solve:

{{{matrix(1,3,x/144,""="",64/x)}}}

{{{matrix(1,3,x^2,""="",9216)}}}

{{{matrix(1,3,x,""="",sqrt(9216))}}}

{{{matrix(1,3,x,""="",96)}}}

So

{{{matrix(1,7,r,""="",x/144,""="",96/144,""="",2/3)}}}

{{{matrix(1,11,
S[infinity],
""="",
a[1]/(1-r),
""="",
144/(1-2/3),
""="",
144/(1/3),
""="",
144*3,
""="",
432)}}}

Edwin</pre>