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In a sequence of integers, each term in the first m integers is 5
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28 Mar 2018, 10:17
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Question Stats:
63% (03:07) correct 37% (02:57) wrong based on 49 sessions
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In a sequence of integers, each term in the first m integers is 5 greater than the previous integer. Thereafter, each integer is 2 less than the previous integer. If the first integer and the last integer in the sequence are each 50 and the number of integers in the sequence is 29, m=?
Re: In a sequence of integers, each term in the first m integers is 5
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28 Mar 2018, 12:52
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pushpitkc wrote:
In a sequence of integers, each term in the first m integers is 5 greater than the previous integer. Thereafter, each integer is 2 less than the previous integer. If the first integer and the last integer in the sequence are each 50 and the number of integers in the sequence is 29, m=?
In a sequence of integers, each term in the first m integers is 5
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Updated on: 25 Apr 2018, 15:33
pushpitkc wrote:
In a sequence of integers, each term in the first m integers is 5 greater than the previous integer. Thereafter, each integer is 2 less than the previous integer. If the first integer and the last integer in the sequence are each 50 and the number of integers in the sequence is 29, m=?
Re: In a sequence of integers, each term in the first m integers is 5
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24 Apr 2018, 00:09
pushpitkc wrote:
In a sequence of integers, each term in the first m integers is 5 greater than the previous integer. Thereafter, each integer is 2 less than the previous integer. If the first integer and the last integer in the sequence are each 50 and the number of integers in the sequence is 29, m=?
Can you please specify which part is troubling you? I have provided with a detailed solution as follows...
Kindly go through it once and let me know. I would be happy to help
Best, Gladi
Gladiator59 wrote:
pushpitkc wrote:
In a sequence of integers, each term in the first m integers is 5 greater than the previous integer. Thereafter, each integer is 2 less than the previous integer. If the first integer and the last integer in the sequence are each 50 and the number of integers in the sequence is 29, m=?
Re: In a sequence of integers, each term in the first m integers is 5
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24 Apr 2018, 00:46
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pushpitkc wrote:
In a sequence of integers, each term in the first m integers is 5 greater than the previous integer. Thereafter, each integer is 2 less than the previous integer. If the first integer and the last integer in the sequence are each 50 and the number of integers in the sequence is 29, m=?
First, we have m integers, for which each term is 5 greater than the previous integer: 50, 55, 60, 65, 70, ..., 50 + 5(m - 1) Notice, that the last integer in this, first, part of the sequence is \(50 + 5(m - 1)\). \(n_{th}\) term in the arithmetic progression (AP) is given by \(a_n=a_1+d(n-1)\), where d is the common difference.
After those m integers, each integer is 2 less than the previous integer: \([50 + 5(m - 1)] - 2\), \([50 + 5(m - 1)] - 4\), \([50 + 5(m - 1)] - 6\), ..., 50. Notice, that since there are a total of 29 integers in the whole sequence, then the above continues for 29 - m integers. The last integer given to be 50, thus by applying the same rule for AP, we'll get \(50 = [50 + 5(m - 1) - 2] + (-2)[(29 - m) - 1]\):
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63% (03:07) correct 
