The squares of 3 consecutive terms of an arithmetic progression are a − 3 3 6 , a and a + 6 2 4 . What is the value of a ?
This problem is proposed by Ahmad .
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let x be the first term of the progression, and y be the common difference. then x²=a-336 (1)
(x+y)²=a (2)
(x+2y)²=a+624 (3) by expanding we have: x²=a-336 (1)
x²+2xy+y²=a (2)
x²+4xy+4y²=624 (3)
by subtracting side by side (1) from (2) we get: 2xy+y²=336 (4). by doing the same for (3) and (1) we get: 4xy+4y²=960 thus 2xy+2y²=480 (5)
subtracting (4) and (5) we get y²=144, then y=12 or y= -12
replacing y in (2) we get
x²-24x+144=a or x²+24x+144=a subtracting with (1) we find that -24x=192 or 24x=192 thus |x|=8, and finally x²=64
finnally, replacing in (1) : a=64+336=440
Let x 1 be the first number, x 2 be the 2nd number, x 3 be the third number of the arithmetic progression.
In an arithmetic progression, difference of two consecutive terms is constant. Thus, d = x 2 − x 1 = x 3 − x 2
x 2 − x 1 = x 3 − x 2
x 2 + x 2 = x 3 + x 1
( 2 x 2 ) 2 = ( x 3 + x 1 ) 2
4 x 2 2 = x 3 2 + 2 x 1 x 3 + x 1 2
since, x 1 2 = a − 3 3 6 ; x 2 2 = a ; x 3 2 = a + 6 2 4
thus, x 1 = a − 3 3 6 ; x 3 = a + 6 2 4
4 x 2 2 = x 3 2 + 2 x 1 x 3 + x 1 2
4 a = ( a + 6 2 4 ) + 2 a − 3 3 6 ⋅ a + 6 2 4 + ( a − 3 3 6 )
4 a = ( 2 a + 2 8 8 ) + 2 a 2 + 2 8 8 a − 2 0 9 6 6 4
2 a − 2 8 8 = 2 a 2 + 2 8 8 a − 2 0 9 6 6 4
2 ( a − 1 4 4 ) = 2 a 2 + 2 8 8 a − 2 0 9 6 6 4
( a − 1 4 4 ) 2 = ( a 2 + 2 8 8 a − 2 0 9 6 6 4 ) 2
a 2 − 2 8 8 a + 2 0 7 3 6 = a 2 + 2 8 8 a − 2 0 9 6 6 4
2 8 8 a + 2 8 8 a = 2 0 7 3 6 + 2 0 9 6 6 4
5 7 6 a = 2 3 0 4 0 0
a = 4 0 0 Ans.
Let the middle term of this A.P. be x . Let the common difference between the terms be y .
Therefore, the terms of this A.P. are x − y , x and x + y .
It is given that a − 3 3 6 , a and a + 6 2 4 are the squares of these terms respectively.
Therefore we have equation 1 (\a - 336 = x^2 - 2xy + y^2), equation 2 a = x 2 and equation 3 a + 6 2 4 = x 2 + 2 x y + y 2 .
Using equation 1 we can cancel out a = x 2 from equations 2 and 3 .
Therefore we are left with 6 2 4 = 2 x y + y 2 and − 3 3 6 = − 2 x y + y 2 .
Adding these two new equations together we can eliminate x and we are left with 2 8 8 = 2 y 2 .
Solving this we get y = 1 2 . Using the y = 1 2 in equation 1 we get x = 2 0 .
Therefore a = x 2 , a = 4 0 0
I have made a formatting error in equation 1 it should mean a − 3 3 6 = x 2 − 2 x y + y 2 .
Also it is a bit confuing in how it appears but I mean to say equation 1: ... equation 2: ... and equation 3:... It is appearing as if I have multiplied by 3 in equation 3 although that is not the case
Suppose that, the 3 consecutive terms terms would be x − y , x , x + y , ( x + y ) 2 = a + 3 3 6 , x 2 = a , ( x + y ) 2 = a + 6 2 4 , x 2 − 2 x y + y 2 = 3 3 6 , x 2 = a , x 2 + 2 x y + y 2 = a + 6 2 4 ) ⇒ − 2 x y + a 2 = − 3 3 6 , 2 x y + a 2 = 6 2 4 , 2 a 2 = 2 8 8 ⇒ a = 1 2 . x = 2 0 . a = 2 0 2 = 4 0 0
Suppose the three terms of the A.P be (b-d),b,(b+d); Hence, as in the question, b^2-(b-d)^2=336; =>2bd-d^2=336 .............eqn. 1; Again, (b+d)^2-b^2=624; =>2bd+d^2=624 ...............eqn 2; Adding equations 1 and 2, we get, 4bd=960; =>bd=240; =>b=240/d; Putting the value of b in the equation 1, we get, 480-d^2=336; =>d^2=144; =>d=12 or d= -12; let us assume d=12; Putting the value of d in equation, 24b-144=336; =>b=20; Needless to say, since a is the square of the middle term, Hence, a=b^2=400.
As we have the squares of 3 consecutive numbers of an Arithmetic Progression as, ( a − 3 3 6 ) , a , ( a + 6 2 4 ) Let's have ( x − d ) , x , ( x + d ) as the three consecutive numbers of an arithmetic progression with d as the common difference.Now we can write the square of the first term as, ( x − d ) 2 = a − 3 3 6 ⇒ x 2 − 2 x d + d 2 = a − 3 3 6 − − − − − ( 1 ) Then we can write the square of the second term as, x 2 = a − − − − − ( 2 ) Then we can write the square of the third term as, ( x + d ) 2 = a + 6 2 4 ⇒ x 2 + 2 x d + d 2 = a + 6 2 4 − − − − − ( 3 ) Subtracting equation (1) from equation (3),we get, 4 x d = 9 6 0 ⇒ x d = 4 9 6 0 ⇒ x d = 2 4 0 − − − − − ( 4 ) Subtracting equation (1) from equation (2), we get 2 x d − d 2 = 3 3 6 2 × 2 4 0 − 3 3 6 = d 2 ⇒ 4 8 0 − 3 3 6 = d 2 ⇒ d 2 = 1 4 4 ⇒ d = 1 2 Substituting the value of d in equation (4),we get, x = 1 2 2 4 0 ⇒ x = 2 0 But we know that, a = x 2 ⇒ a = 2 0 2 ⇒ a = 4 0 0
Let the 3 consecutive terms of the arithmetic sequence be ( b − k ) , b , ( b + k ) . Then, their squares would just be ( b 2 − 2 k b + k 2 ) = a − 3 3 6 , b 2 = a , b 2 + 2 k b + k 2 = a + 6 2 4 .
From this, it is clear that 4 k b = 6 2 4 + 3 3 6 = 9 6 0 and that 2 k b + k 2 = 6 2 4 . Then, we can know that k 2 = 6 2 4 − 2 9 6 0 = 1 4 4 , hence k = 1 2 . Substituting the value of k , we can get b = 4 × 1 2 9 6 0 = 2 0 . Here, we arrive at our last step: a = b 2 = 4 0 0 .
Let the consecutive terms of the A.P be x , x + d , x + 2 d
Thus x 2 = a − 3 3 6 , ( x + d ) 2 = a and ( x + 2 d ) 2 = a + 6 2 4
i.e x 2 = a − 3 3 6 . . . . . ( 1 ) , x 2 + 2 d x + d 2 = a . . . . . ( 2 ) and x 2 + 4 x d + 4 d 2 = a + 6 2 4 . . . . . ( 3 )
From.....(1) substitute (a-336) into .....(2) and .....(3)
i.e. ( a − 3 3 6 ) + 2 d x + d 2 = a
from which we get d 2 + 2 d x = 3 3 6 . . . . . ( 4 )
and ( a − 3 3 6 ) + 4 x d + 4 d 2 = a + 6 2 4
from which we get 4 d 2 + 4 d x = 9 6 0 . . . . . ( 5 )
solving .....(4) and .....(5) simultaneously, we get d = 1 2 and x = 8
but a = ( x + d ) 2
i.e. a = ( 8 + 1 2 ) 2
Therefore a = ( 2 0 ) 2 = 4 0 0
We have that a − 3 3 6 , a and a + 6 2 4 form an arithmetic progression. Therefore
2 a = a − 3 3 6 + a + 6 2 4
4 a = 2 a + 2 8 8 + 2 ( a − 3 3 6 ) ( a + 6 2 4 )
a − 1 4 4 = ( a − 3 3 6 ) ( a + 6 2 4 )
a 2 − 2 8 8 a + 2 0 7 3 6 = a 2 + 2 8 8 a − 2 0 9 6 6 4
5 7 6 a = 2 3 0 4 0 0
a = 4 0 0
why is root(a) multiplied by 2 in the solution
let the common difference of the a.p. be 'd' now (x-d)(x-d)=a-336...(1) x.x=a.................(2) (x+d)(x+d)=a+624.(3)
4xd=960;2xd=480;2xd+d.d=624;480+d.d=624;d=12
x=20;d=400
(middle term-d)2=a-336 middle term squared =a (middle term +common difference )squared =a+624 a=400
i have just used simple arithmatic . As all three numbers are squares so a will defenately be greater than 0 and thus it must has to be greater than 336 . If we square root 336 we will get 18.33 so root of a can't be 18 . If we try 19 it also don't follow the sequence but if 20 be the root if a , it follows the sequance and so a will be the square of 20 which means it must be 400 .
Let x be the second number of the 3 consecutive terms.
Let y be the difference of the consecutive terms.
System of Equations:
1) (x-y)^{2}=a-336
2) x^{2}=a
3) (x+y)^{2}=a+624
Solution:
Sub 2) into 1)
(x-y)^{2}=x^{2}-336
4) 2xy=336+y^{2}
Sub 2) into 3)
(x+y)^{2}=x^{2}+624
5) 2xy=624-y^{2}
Elimination: Subtract 5) from 4)
0=-288+2y^{2}
y=12
Sub y=12 into 4)
2x(12)=336+12^{2}
x=20
Sub x=20 into 2)
a=400
b = first term of the arithmetic progression
r = reason of arithmetic progression
b² = a-336
b² + 2br + r² = a
b²+4br+4r² = a + 624
Replacing b in the equations
2br+r²=336 => -4br + r² = 336 = II
4br+4r²=960 => 4br+4r²=960 = III
II+II = 2r²=288 > r² = 144 => r = 12
Replacing r in equation
2br+ r² = 336
2×12×b+144=336
24×b=192
b=8
Replacing b in equation
b = a-336
64= a-336
a = 400
Because these are the square terms of an arithmetic sequence we know that...
√a - √(a – 336) = √(a + 624) - √a
(2√a)^2 = ((√(a + 624)) + (√(a – 336)))^2
4a = (a + 624) + (2√(a – 336) * √(a + 624)) + (a - 336)
4a = 2a + 288 + (2√(a – 336) * √(a + 624))
2a - 288 = (2√(a – 336) * √(a + 624))
a - 144 = √(a – 336) * √(a + 624)
(a - 144)^2 = (a + 624)(a - 336)
a^2 - 288a + 20736 = a^2 + 288a - 209664
-288a = 288a - 230400
-576a = -230400
a = 400
So...
64, 400, 1024 are the squares of 8, 20, and 32 respectively... and they are each 12 apart.
An arithmetic progression has a common difference, d from each term. Therefore, the terms a - 336 , a , a + 624 can be represented as ( b − d ) 2 , b 2 , ( b + d ) 2 .
After expansion from the new representation, we obtain b 2 - 2bd + d 2 , b 2 , b 2 + 2bd + d 2 . Notice that the difference between terms are 2bd + d 2 .
By comparing the similar progression: a - 336 , a , a + 624 and b 2 - 2bd + d 2 , b 2 , b 2 + 2bd + d 2 , we take the differences and form two equations;
Eliminating the term ' 2bd ' by equation 1+2, we obtain d = 12 . Substituting d = 12 into equation 1 or 2 obtains b = 20 . Hence, b 2 = a = 2 0 2 = 4 0 0
a = x 2
Therefore, a − 3 3 6 , a , a + 6 2 4 = x − d , x , x + d
x 2 − ( x − d ) 2 = 3 3 6
x 2 − x 2 − d 2 + 2 d x = 3 3 6
− d 2 + 2 d x = 3 3 6
( x + d ) 2 − x 2 = 6 2 4
x 2 + d 2 + 2 d x − x 2 = 6 2 4
d 2 + 2 d x = 6 2 4
2 d 2 = 2 8 8
d = 1 2
1 4 4 + 2 4 x = 6 2 4
2 4 x = 4 8 0
x = 2 0
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An arithmetic progression is a sequence of numbers such that the difference between the consecutive terms is constant. We are going to start by assuming that the 3 consecutive terms of the given progression are ( k − d ) , k and ( k + d ) where d is the common difference between the terms. According to the problem, ( k − d ) 2 = a − 3 3 6 . . . ( 1 )
k 2 = a . . . ( 2 ) and
( k + d ) 2 = a + 6 2 4 . . . ( 3 ) .
If we add ( 1 ) and ( 3 ) we get something like this: 2 ( k 2 + d 2 ) = 2 a + 2 8 8
⇒ ( k 2 + d 2 ) = a + 1 4 4 .
And since k 2 = a [see ( 2 ) ], this can be re-written as d 2 = 1 4 4 which tells us d is either 1 2 or − 1 2 .
Now if we subtract ( 1 ) from ( 3 ) , we get 4 k d = 9 6 0 ⇒ k d = 2 4 0 . This implies, k = 2 0 when d = 1 2 and k = − 2 0 when d = − 1 2 . We have to find a which is equal to k 2 . Therefore, a = ( 2 0 ) 2 = ( − 2 0 ) 2 = 4 0 0 .