SOLUTION: find the sum of all the integers between 100 and 500 and are divisble by 8 (B) the first 3terms in a G.P are 144,x and 64 where x is a positive find the sum of infinity of a progr

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:

a1 = 104 and an = 496








The formula for the sum is









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We solve:









So





Edwin