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24 Oct 2023, 03:38
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A
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Difficulty:
55%
(hard)
Question Stats:
66% (02:30) correct
34%
(02:52)
wrong
based on 93
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If a positive integer ‘n’ is subtracted from the squares of three consecutive terms of an Arithmetic Progression, the numbers obtained are 108, 220 and 364 respectively. What is the sum of the digits of ‘n’?
A. 9
B. 8
C. 7
D. 6
E. 4
A. 9
B. 8
C. 7
D. 6
E. 4
24 Oct 2023, 03:56
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I went through a rather long approach here. If anyone can suggest an easier approach, that would be great!
Here's my approach:
Let the 3 consecutive integers in the Arithmetic progression be (a-d), (a) and (a+d) respectively.
Now according to the question stem,
(a-d)² - n = 108 => (a-d)² = 108 +n ------->Eq.1
(a)² - n = 220 => (a)² = 220 +n ------->Eq.2
(a+d)² - n = 364 => (a+d)² = 364 +n ------->Eq.3
Subtracting equation 2 from equation 3 we get 2ad+d² = 144 -----> Eq.4
Subtracting equation 2 from equation 1 we get d²-2ad = -112 -----> Eq.5
Adding equations 4 and 5, we get d²=16 => d =4
Substituting the value of d in either of the equations 4 and 5, we get the value of a
d²-2ad = -112 => 16-2*a*4 = -112 which gives 128 = 8a and in turn a = 16
So now we know a = 16 and d = 4
The AP will be (16-4), 16, (16+4) ie the AP will be 12,16,20...
The condition mentioned in the question stem is that n is a positive integer subtracted from the squares of the terms in the AP
From equation 1, (a-d)² - n = 108 => 12² - n =108 => 144 - n = 108 hence n is 36.
Now we know that the positive integer n is 36.
The sum of the digits of n is 3+6 = 9
The answer is A
Here's my approach:
Let the 3 consecutive integers in the Arithmetic progression be (a-d), (a) and (a+d) respectively.
Now according to the question stem,
(a-d)² - n = 108 => (a-d)² = 108 +n ------->Eq.1
(a)² - n = 220 => (a)² = 220 +n ------->Eq.2
(a+d)² - n = 364 => (a+d)² = 364 +n ------->Eq.3
Subtracting equation 2 from equation 3 we get 2ad+d² = 144 -----> Eq.4
Subtracting equation 2 from equation 1 we get d²-2ad = -112 -----> Eq.5
Adding equations 4 and 5, we get d²=16 => d =4
Substituting the value of d in either of the equations 4 and 5, we get the value of a
d²-2ad = -112 => 16-2*a*4 = -112 which gives 128 = 8a and in turn a = 16
So now we know a = 16 and d = 4
The AP will be (16-4), 16, (16+4) ie the AP will be 12,16,20...
The condition mentioned in the question stem is that n is a positive integer subtracted from the squares of the terms in the AP
From equation 1, (a-d)² - n = 108 => 12² - n =108 => 144 - n = 108 hence n is 36.
Now we know that the positive integer n is 36.
The sum of the digits of n is 3+6 = 9
The answer is A
01 Nov 2023, 01:34
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athena97
Instead of using algebra, it would be simpler to just solve it normally?
Next square to 108 is 121 (\(11^2\)), difference is 13.
Adding 13 to 220 does not give a square.
Then the next one is 144 (\(12^2\)), difference from 108 is 36.
Adding 36 to 220 gives us a square (256, \(16^2\))
Adding 36 to 364 also gives us a square (400, \(20^2\))
Hence 36 is the integer n, and the sum 9
