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..
Number Systems (Arthimetic Progression)
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The sum of the series is 5^2+6^2+...+44^2+45×20, or (44×45×89/6)-30+900.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..
Therefore answer is d.
Sorry, I can't speak English well. If you can't understand this, please ask me.
- Maciek
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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 = 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
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:
Sum of 44 consecutive squares:S = ( ( n * ( n + 1 ) * ( 2n + 1 ) ) / 6)
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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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 thisMaciek 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:Sum of 44 consecutive squares:S = ( ( n * ( n + 1 ) * ( 2n + 1 ) ) / 6)
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
- Maciek
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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
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
if you have any questions, send me a private message!
should you find this post useful, please click on "thanks" button