SOLUTION: what is the sum of the integers between 20 and 480 that are multiples of 11? please use the equation needed, thank you:)

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Question 275375: what is the sum of the integers between 20 and 480 that are multiples of 11?
please use the equation needed, thank you:)

Found 2 solutions by user_dude2008, Edwin McCravy:
Answer by user_dude2008(1862) About Me  (Show Source):
You can put this solution on YOUR website!
22+33+44+...+451+462+473 = 11(2+3+4+...+41+42+43)


11(2+3+4+...+41+42+43) = 11(1+2+3+4+...+41+42+43-1)

11(2+3+4+...+41+42+43) = 11((43(43+1))/2-1)

11(2+3+4+...+41+42+43) = 11(946-1)

11(2+3+4+...+41+42+43) = 11(945)

11(2+3+4+...+41+42+43) = 10395


Answer: 22+33+44+...+451+462+473 = 10395

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
what is the sum of the integers between 20 and 480 that are multiples of 11?
please use the equation needed, thank you:)

The first term is 22 and we can find the last term by dividing

480 by 11, getting 43%267%2F11 So 43%2A11 or 473 is the
last term.

The multiples of 11 are 11 apart, so we have an arithmetic
sequence for which the common difference, d is 11.

The formula for the sum which we need is this:

S%5Bn%5D=%28n%2F2%29%28a%5B1%5D%2Ba%5Bn%5D%29

where a%5B1%5D=22 is the first term and a%5Bn%5D=473 is the last
term.

S%5Bn%5D=%28n%2F2%29%28a%5B1%5D%2Ba%5Bn%5D%29

But we need n, so we first need the formula for the nth term:

a%5Bn%5D=a%5B1%5D%2B%28n-1%29d

473=22%2B%28n-1%29%2A11

473=22%2B11%28n-1%29

473=22%2B11n-11

473=11%2B11n

462=11n

42=n

No we can substitute 

S%5Bn%5D=%28n%2F2%29%28a%5B1%5D%2Ba%5Bn%5D%29

where n=42, a%5B1%5D=22 and a%5Bn%5D=a%5B22%5D=473 

S%5B42%5D=%2842%2F2%29%2822%2B473%29

S%5B42%5D=%2821%29%28495%29

S%5B42%5D=10395

Edwin