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08 Sep 2020, 22:59
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A
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Difficulty:
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72% (02:24) correct
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The sum of three numbers in an arithmetic progression is 36. The sum of the squares of the three numbers is 464. Find the smallest number?
A.8
B.10
C.12
D.16
E.13
A.8
B.10
C.12
D.16
E.13
08 Sep 2020, 23:35
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Explanation:
let first term = a-d
2nd will be = a
3rd will be = a+d
sum = 36
a-d+a+a+d = 36
3a=36
a=12
sum of the squares = 464
(a-d)^2 + a^2 + (a+d)^2 = 464
3a^2 + 2d^2 = 464
2d^2 = 464 - 3x(144)
d^2 = 32/2
d= 4
Terms : 8,12,16
Smallest is 8
IMO-A
let first term = a-d
2nd will be = a
3rd will be = a+d
sum = 36
a-d+a+a+d = 36
3a=36
a=12
sum of the squares = 464
(a-d)^2 + a^2 + (a+d)^2 = 464
3a^2 + 2d^2 = 464
2d^2 = 464 - 3x(144)
d^2 = 32/2
d= 4
Terms : 8,12,16
Smallest is 8
IMO-A
General Discussion
09 Sep 2020, 10:08
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Yogananda
You don't need to know the meaning of the phrase "arithmetic progression" on the GMAT - what is the source? An arithmetic sequence or progression is one that is equally spaced.
The average of an equally spaced list is always equal to its median. So if 3 equally spaced numbers sum to 36, and thus average to 12, then 12 is the median. The smallest number in the list must be less than the median, so only A or B could be right. The units digit of 10^2 + 12^2 + 14^2 will be zero (since it must be the units digit of 0+4+6), so B cannot be right, and the answer must be A.
I suppose you might want to consider the possibility the list is 12, 12, 12, but adding the squares of those numbers must give us a multiple of 3, and 464 is not a multiple of 3 (or you could use units digits).
Alternatively, you can solve exactly: since 12 is our median, our list must be 12-d, 12, 12+d. Squaring each term and adding, we get 464, so
(12-d)^2 + 12^2 + (12 + d)^2 = 464
(12 - d)^2 + (12 + d)^2 = 320
144 - 24d + d^2 + 144 + 24d + d^2 = 320
2d^2 = 32
d^2 = 16
d = 4 or -4
and either way, the sequence is 8, 12, 16.
