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For every positive integer n, the nth term of a sequence is
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Updated on: 09 Nov 2012, 02:30
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For every positive integer n, the nth term of a sequence is the total of three consecutive integers starting at n. What is the total of terms 1 through 99 of this series? A. 5,250 B. 10,098 C. 14,850 D. 15,147 E. 15,150 Average of sequence is average of first and last terms, or (6+300)/2 = 306/2 = 153.
Number of terms is 99, so the sum of the series is 153*99 = 15,147 (subtract 153 from 153*100 = 15,300)
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Originally posted by anon1 on 08 Nov 2012, 17:41.
Last edited by Bunuel on 09 Nov 2012, 02:30, edited 1 time in total.
Edited the question.




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Re: For every positive integer n, the nth term of a sequence is
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09 Nov 2012, 02:39
anon1 wrote: For every positive integer n, the nth term of a sequence is the total of three consecutive integers starting at n. What is the total of terms 1 through 99 of this series? A. 5,250 B. 10,098 C. 14,850 D. 15,147 E. 15,150 Average of sequence is average of first and last terms, or (6+300)/2 = 306/2 = 153.
Number of terms is 99, so the sum of the series is 153*99 = 15,147 (subtract 153 from 153*100 = 15,300) According to the stem n th term of the sequence equals to n+(n+1)+(n+2)=3(n+1). Thus: The first term equals to 3(1+1)=6; The second term equals to 3(2+1)=9; The third term equals to 3(3+1)=12; ... The 99 th term equals to 3(99+1)=300. As we can see we have an evenly spaced set. The sum of the terms in any evenly spaced set is (average)*(# of terms)=(6+300)/2*99=153*99=15,147. Answer: D.
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Re: For every positive integer n, the nth term of a sequence is
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08 Nov 2012, 21:32
anon1 wrote: For every positive integer n, the nth term of a sequence is the total of three consecutive integers starting at n. What is the total of terms 1 through 99 of this series? 5,250 10,098 14,850 15,147 15,150 Could someone help explain what this question is asking? Average of sequence is average of first and last terms, or (6+300)/2 = 306/2 = 153. Number of terms is 99, so the sum of the series is 153*99 = 15,147 (subtract 153 from 153*100 = 15,300) question defines a series, \(S_n = n+ (n+1) + n(+2)=3(n+1)\) and asking sum of \(S = S_1 to S_{99}\) Now since \(S_n= 3(n+1)\) \(S = 3 *[2+3+.....+100]\) \(S= 3 *[1+2+3+....+100]  3\) \(S= 3*100*101/2 3\) \(S=15147\) Ans D it is. Other method is, since \(S_n= 3(n+1)\) First term, \(S_1 = 6\) Last term, \(S_{99}=300\) total number of terms = 1 to 99 = 99 Sum of all terms = Average * Number of terms =306/2 * 99 =153*99 A number that must end with 7 So Ans D it is.
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Re: For every positive integer n, the nth term of a sequence is
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08 Nov 2012, 21:43
The nth term of the sequence is the sum of three consecutive integers starting with n => T(n) = n + (n+1) + (n+2) => T(n) = 3(n+1) Therefore S(n) = 3*[(n(n+1)/2 + n] = (3/2)*[n(n+3)] => S(99) = (3/2)*[99*102] = 3*99*51 = 15,147 Option (D)
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Re: For every positive integer n, the nth term of a sequence is
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08 Nov 2012, 22:37
Ahh, I got it now. It made sense after I plugged in with a sample number of n=5. When I was doing it at first the wording was so confusing. Plugging in to make one example really helps to translate words into math.



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Re: For every positive integer n, the nth term of a sequence is
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08 Nov 2012, 22:40
What helped me the most was seeing exactly what kind of set it was.
The minute I saw it was a set beginning with 6 with an interval of 3 and ending at the 99th place it became a matter of the multiplying the average times the # of numbers. which is easy breezy (153x99)



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Re: For every positive integer n, the nth term of a sequence is
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22 Sep 2016, 22:03
anon1 wrote: For every positive integer n, the nth term of a sequence is the total of three consecutive integers starting at n. What is the total of terms 1 through 99 of this series? A. 5,250 B. 10,098 C. 14,850 D. 15,147 E. 15,150 Average of sequence is average of first and last terms, or (6+300)/2 = 306/2 = 153.
Number of terms is 99, so the sum of the series is 153*99 = 15,147 (subtract 153 from 153*100 = 15,300) We know that each term will be the sum of = n+ n+1 + n+2 = 3(n+1) Therefore for sum of series will be = 3(2+3+4...100) > divisible by 3. Search the options which are divisible by 3. Option D fits the bill. (sum of all digits is divisible by 3)



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Re: For every positive integer n, the nth term of a sequence is
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23 Sep 2016, 02:46
anon1 wrote: For every positive integer n, the nth term of a sequence is the total of three consecutive integers starting at n. What is the total of terms 1 through 99 of this series? A. 5,250 B. 10,098 C. 14,850 D. 15,147 E. 15,150 Average of sequence is average of first and last terms, or (6+300)/2 = 306/2 = 153.
Number of terms is 99, so the sum of the series is 153*99 = 15,147 (subtract 153 from 153*100 = 15,300) \(t_n = n + (n+1) + (n+2)\) \(S(99) = t_1 + t_2 + ... + t_{99}\) S(99) = ( 1+2+3) + ( 2+3+4) + ( 3+4+5) + ... + ( 99+100+101) This can be taken to be sum of 3 series \((1 + 2 + 3 + ... + 99) + (2 + 3 + ... 100) + (3 + 4 + 5 + ... + 101)\) \(Sum = 99*100/2 + (100*101/2  1) + (101*102/2  1  2)\) \(Sum = 99*50 + 101*(50 + 51)  4 =4950 + 101^2  4\) This will end with 7 (because of 1  4) so answer will be (D)
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Re: For every positive integer n, the nth term of a sequence is
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26 Jul 2017, 06:24
anon1 wrote: For every positive integer n, the nth term of a sequence is the total of three consecutive integers starting at n. What is the total of terms 1 through 99 of this series? A. 5,250 B. 10,098 C. 14,850 D. 15,147 E. 15,150 Average of sequence is average of first and last terms, or (6+300)/2 = 306/2 = 153.
Number of terms is 99, so the sum of the series is 153*99 = 15,147 (subtract 153 from 153*100 = 15,300) tn = n+(n+1)+(n+2) = 3n +3 t1 + t2 + t3 +t4 +t5 + t6 ...... + t99 = Sum (1,99) of 3n + 3 = [3n(n+1)2 +3n ] (1,99) = 3*99*100/2 + 3 *99  0 = 3*(99*51) = 15147 Answer C
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