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Number Systems (Arthimetic Progression)

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by kapur.arnav » Sun Sep 05, 2010 9:35 am
Consider the following series: 5,5,5,5,5,6,6,6,6,6,6,7,7,7,7,7,7,7.... and so on.. what is the sum of the first 1000 terms of the series?

a. 33295
b. 30270
c. 32265
d. 30240
e. 30965

Please explain the approach.. thanks..
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by zino0067 » Mon Sep 06, 2010 1:40 am
kapur.arnav wrote:Consider the following series: 5,5,5,5,5,6,6,6,6,6,6,7,7,7,7,7,7,7.... and so on.. what is the sum of the first 1000 terms of the series?

a. 33295
b. 30270
c. 32265
d. 30240
e. 30965

Please explain the approach.. thanks..
The sum of the series is 5^2+6^2+...+44^2+45×20, or (44×45×89/6)-30+900.
Therefore answer is d.

Sorry, I can't speak English well. If you can't understand this, please ask me.
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by Maciek » Mon Sep 06, 2010 10:21 am
Hi,

Your solution is excellent!

Let Z be the sum of the first 1000 terms of our series

if n = 1000 then an = ?
5, 6, 7, 8, 9, ..., 44
sum1 = 0.5*40(5 + 44) = 20*49 = 980
45 will be x times
x = 1000 - 980 = 20

The formula for sum of consecutive squares:
S = ( ( n * ( n + 1 ) * ( 2n + 1 ) ) / 6)
Sum of 44 consecutive squares:
S = 1^2 + 2^2 +3^2 + 4^2 + 5^2 + 6^2 + ... + 44^2 = (44*(44 + 1)*(2*44 + 1))/6

Z = 5^2 + 6^2 + ... + 44^2 + 45*20
S = 1^2 + 2^2 + 3^2 + 4^2 + (Z - 45*20)
Z = S + 45*20 - 1 - 4 - 9 - 16
Z = ((44*(44 + 1)*(2*44 + 1))/6) + 900 - 30
Z = (44*45*89)/6 + 870
Z = 22*15*89 + 870
Z = 30240

Hope it helps!
Best,
Maciek
"There is no greater wealth in a nation than that of being made up of learned citizens." Pope John Paul II

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by narik11 » Thu Sep 16, 2010 9:46 am
Maciek wrote:Hi,

Your solution is excellent!

Let Z be the sum of the first 1000 terms of our series

if n = 1000 then an = ?
5, 6, 7, 8, 9, ..., 44
sum1 = 0.5*40(5 + 44) = 20*49 = 980
45 will be x times
x = 1000 - 980 = 20

The formula for sum of consecutive squares:
S = ( ( n * ( n + 1 ) * ( 2n + 1 ) ) / 6)
Sum of 44 consecutive squares:
S = 1^2 + 2^2 +3^2 + 4^2 + 5^2 + 6^2 + ... + 44^2 = (44*(44 + 1)*(2*44 + 1))/6

Z = 5^2 + 6^2 + ... + 44^2 + 45*20
S = 1^2 + 2^2 + 3^2 + 4^2 + (Z - 45*20)
Z = S + 45*20 - 1 - 4 - 9 - 16
Z = ((44*(44 + 1)*(2*44 + 1))/6) + 900 - 30
Z = (44*45*89)/6 + 870
Z = 22*15*89 + 870
Z = 30240

Hope it helps!
Best,
Maciek
Hi I am able to get the calculation part. But i am not getting how you are interpreting 1000 numbers as 5^2 to 44^2. can u explain on this
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by Maciek » Thu Sep 16, 2010 10:25 am
Hi!
I hope I understand your question correctly.
Look at our series:
5,5,5,5,5,6,6,6,6,6,6,7,7,7,7,7,7,7.... an
There are 1000 terms in the series.
the 5 first terms all are equal to 5
the 6 next terms all are equal to 6
the 7 next terms all are equal to 7

we are looking for an:
5 - 14
5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14= 95
15 - 24
10*10 + 95 = 195
25 - 34
10*10 + 195 = 295
35 - 44
10*10 + 295 = 395
stop! We have found 980 terms, so the last 20 terms must be equal to 45
an = 45

Hope it helps!
Best,
Maciek
"There is no greater wealth in a nation than that of being made up of learned citizens." Pope John Paul II

if you have any questions, send me a private message!

should you find this post useful, please click on "thanks" button :)
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