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Re: A sequence consists of N consecutive even integers in descending order
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01 Jun 2020, 10:53
Since set consist of even consecutive no.
Set will be {N,N-2,N-4,..............,(-N+2)}
Let's solve by options
1) N = 600.
Set will be (600,598,596,594,..........upto 600 term).
The average of first 5 no. Is (600+598+596+594+592)/5=596
And last no. Is -598
[ An = A+(n-1)D
A600=600+(600-1)-2
A600=600+(599)*-2
A600=600-1198
A600=-598 ]
The difference between the average and last no is 596-(-598) => 1194≠1594 (option A rejected)
2) N = 750.
Set will be (750,748,746,744,..........upto 750 term).
The average of first 5 no. Is (750+748+746+744+742)/5=746
And last no. Is -748
[ An = A+(n-1)D
A750=750+(750-1)-2
A750=750+(749)*-2
A750=750-1498
A750=-748 ]
The difference between the average and last no is 746-(-748) => 1494≠1594 (option B rejected)
3) N = 700.
Set will be (700,698,696,694,..........upto 700 term).
The average of first 5 no. Is (700+698+696+694+692)/5=696
And last no. Is -698
[ An = A+(n-1)D
A700=700+(700-1)-2
A700=700+(699)*-2
A700=700-1398
A700=-698 ]
The difference between the average and last no is 696-(-698) => 1394=1594 (option C rejected)
4) N = 800.
Set will be (800,798,796,794,..........upto 800 term).
The average of first 5 no. Is (800+798+796+794+792)/5=796
And last no. Is -798
[ An = A+(n-1)D
A800=800+(800-1)-2
A800=800+(799)*-2
A800=800-1598
A800=-798 ]
The difference between the average and last no is 796-(-798) => 1594=1594 (option D is qualifying)
5) N = 840.
Set will be (840,838,836,834,..........upto 840 term).
The average of first 5 no. Is (840+838+836+834+832)/5=836
And last no. Is -838
[ An = A+(n-1)D
A840=840+(840-1)-2
A840=840+(839)*-2
A840=840-1678
A840=-838 ]
The difference between the average and last no is 836-(-838) => 1674≠1594 (option E rejected)
Answer is D
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