SOLUTION: An arithmetic progression P consists of n terms. From the progression three different progressions P1, P2 and P3 are created such that P1 is obtained by the 1st, 4th, 7th .... term

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Question 935076: An arithmetic progression P consists of n terms. From the progression three different progressions P1, P2 and P3 are created such that P1 is obtained by the 1st, 4th, 7th .... terms of P, P2 has the 2nd,5th, 8th .....terms of Pand P3 has 3rd, 6th, 9th ..... terms of P. It is found that of P1, P2 and P3 two progressions have the property that their average is itself a term of the original progression P. Which of the following can be a possible value of n?
(a.) 20
(b.) 26
(c.) 36
(d.) Both 1 and 2

For the above question, if the Common difference between the terms of P1 is 6, what is the common difference of P?
(a.) 2
(b.) 3
(c.) 6
(d.) Cannot be determined
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Let the terms of P be a%5B1%5D , a%5B2%5D , ... , with a%5Bh%2B1%5D=a%5Bh%5D%2Bd , d being the common difference between the terms od P .

1) The average of k terms of an arithmetic progression is term number %28k%2B1%29%2F2 of that arithmetic progression if k is odd.
If k%7D%7D+is+even%2C+the+average+of+%7B%7B%7Bk terms is the average of two consecutive terms in the middle of that progression (terms number k%2F2 and k%2F2%2B1 ), and that is not a term of the arithmetic progression.
If P%5B1%5D has an odd number of terms, the average of all the terms of P%5B1%5D will be a term of P%5B1%5D and a term of P .
If P%5B1%5D has an even number of term its average will be the average of two consecutive terms of P%5B1%5D that would be terms a%5Bh%5D and a%5Bh%2B3%5D of P .
In P , after a%5Bh%5D comes a%5Bh%2B1%5D=a%5Bh%5D%2Bd , a%5Bh%2B2%5D=a%5Bh%5D%2B2d , and {{a[h+3]=a[h]+3d}}} . The average of a%5Bh%5D and a%5Bh%2B3%5D would be
.
That is a number between a%5Bh%2B1%5D and a%5Bh%2B2%5D , and is not a term of P .
In sum, if P%5B1%5D has an odd number of terms, the average of its terms is a term of P%5B1%5D and a term of P ; if P%5B1%5D has an even number of terms, the average of its terms is neither a term of P%5B1%5D nor a term of P .
The same can be said of P%5B2%5D and P%5B3%5D .

If P has n=36=3%2A12 terms, P%5B3%5D will have 12 terms,
starting with term number 3=3%2A1 of P , and ending with term number 36=3%2A12 of P . P%5B2%5D and P%5B1%5D will also have 12 terms, because P%5B2%5D starts and ends 1 terms of P before P%5B3%5D , and P%5B1%5D starts and ends 2 terms of P before P%5B3%5D . In that case, all four arithmetic progressions will have an even number of terms, and will have averages that are not terms of P .

If nwere 1 less, P%5B3%5D would have 1 less, but P%5B2%5D and P%5B1%5D would still have 12 terms. That happens for all n that are 1 short of a multiple of 3 , such as
n=3%2A7-1=21-1=20 and n=3%2A9-1=27-1=26 .
For n=3%2A7-1=21-1=20 , P%5B3%5D has 6 terms, but P%5B2%5D and P%5B1%5D have 7 terms. The averages for P%5B2%5D and P%5B1%5D are their respective 4th terms, that are terms of P.
The same thing happens for n=3%2A9-1=27-1=26 , which gives P%5B2%5D and P%5B1%5D 9 terms and their averages are their respective 5th terms.

So, the answer is (d.) Both 1 and 2.

2) The common difference for terms of P%5B1%5D , P%5B2%5D and P%5B1%5D is the same. It is the difference between terms a%5Bh%5D and a%5Bh%2B3%5D of P%7D%7D+.%0D%0A%7B%7B%7Ba%5Bh%2B3%5D-a%5Bh%5D=a%5Bh%5D%2B3d-a%5Bh%5D=3d .
If that is 6 ,
3d=6--->d=6%2F3--->highlight%28d=2%29