Ahmad's sequence of squares

Algebra Level 3

The squares of 3 consecutive terms of an arithmetic progression are a 336 a-336 , a a and a + 624. a + 624. What is the value of a a ?

This problem is proposed by Ahmad .


The answer is 400.

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18 solutions

Mursalin Habib
May 20, 2014

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 3 consecutive terms of the given progression are ( k d ) (k-d) , k k and ( k + d ) (k+d) where d d is the common difference between the terms. According to the problem, ( k d ) 2 = a 336... ( 1 ) (k-d)^2=a-336 . . .(1)
k 2 = a . . . ( 2 ) k^2=a . . . (2) and
( k + d ) 2 = a + 624... ( 3 ) (k+d)^2=a+624 . . .(3) .
If we add ( 1 ) (1) and ( 3 ) (3) we get something like this: 2 ( k 2 + d 2 ) = 2 a + 288 2(k^2+d^2)=2a+288
( k 2 + d 2 ) = a + 144 \Rightarrow (k^2+d^2)=a+144 .
And since k 2 = a k^2=a [see ( 2 ) (2) ], this can be re-written as d 2 = 144 d^2=144 which tells us d d is either 12 12 or 12 -12 .
Now if we subtract ( 1 ) (1) from ( 3 ) (3) , we get 4 k d = 960 4kd=960 k d = 240 \Rightarrow kd=240 . This implies, k = 20 k=20 when d = 12 d=12 and k = 20 k=-20 when d = 12 d=-12 . We have to find a a which is equal to k 2 k^2 . Therefore, a = ( 20 ) 2 = ( 20 ) 2 = 400 a=(20)^2=(-20)^2=400 .


Most submitted correct solutions were similar to this one. One has to be careful that d could be 12 or -12.

Calvin Lin Staff - 7 years ago
Anas Elidrissi
May 20, 2014

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

440 at the end is a clear misprint

Calvin Lin Staff - 7 years ago

Let x 1 x_1 be the first number, x 2 x_2 be the 2nd number, x 3 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 d = x_2-x_1 = x_3-x_2

x 2 x 1 = x 3 x 2 x_2-x_1 = x_3-x_2

x 2 + x 2 = x 3 + x 1 x_2+x_2 = x_3+x_1

( 2 x 2 ) 2 = ( x 3 + x 1 ) 2 {(2x_2)}^2 = {(x_3+x_1)}^2

4 x 2 2 = x 3 2 + 2 x 1 x 3 + x 1 2 4{x_2}^2 = {x_3}^2+2x_1x_3+{x_1}^2

since, x 1 2 = a 336 ; x 2 2 = a ; x 3 2 = a + 624 {x_1}^2 = a-336; {x_2}^2 = a; {x_3}^2 = a+624

thus, x 1 = a 336 ; x 3 = a + 624 x_1 = \sqrt{a-336}; x_3 = \sqrt{a+624}

4 x 2 2 = x 3 2 + 2 x 1 x 3 + x 1 2 4{x_2}^2 = {x_3}^2+2x_1x_3+{x_1}^2

4 a = ( a + 624 ) + 2 a 336 a + 624 + ( a 336 ) 4a = (a+624)+2{\sqrt{a-336}}\cdot{\sqrt{a+624}}+(a-336)

4 a = ( 2 a + 288 ) + 2 a 2 + 288 a 209664 4a = (2a+288)+2{\sqrt{a^2+288a-209664}}

2 a 288 = 2 a 2 + 288 a 209664 2a-288 = 2{\sqrt{a^2+288a-209664}}

2 ( a 144 ) = 2 a 2 + 288 a 209664 2(a-144) = 2{\sqrt{a^2+288a-209664}}

( a 144 ) 2 = ( a 2 + 288 a 209664 ) 2 ({a-144})^2 = ({\sqrt{a^2+288a-209664}})^2

a 2 288 a + 20736 = a 2 + 288 a 209664 {a^2-288a+20736} = {a^2+288a-209664}

288 a + 288 a = 20736 + 209664 {288a+288a} = {20736+209664}

576 a = 230400 {576a} = {230400}

a = 400 {a} = {400} Ans.

A roundabout way, but interesting

Calvin Lin Staff - 7 years ago

Let the middle term of this A.P. be x x . Let the common difference between the terms be y y .

Therefore, the terms of this A.P. are x y x - y , x x and x + y x + y .

It is given that a 336 a - 336 , a a and a + 624 a + 624 are the squares of these terms respectively.

Therefore we have equation 1 1 (\a - 336 = x^2 - 2xy + y^2), equation 2 2 a = x 2 a = x^2 and equation 3 3 a + 624 = x 2 + 2 x y + y 2 a + 624 = x^2 + 2xy + y^2 .

Using equation 1 1 we can cancel out a = x 2 a = x^2 from equations 2 2 and 3 3 .

Therefore we are left with 624 = 2 x y + y 2 624 = 2xy + y^2 and 336 = 2 x y + y 2 -336 = -2xy + y^2 .

Adding these two new equations together we can eliminate x x and we are left with 288 = 2 y 2 288 = 2y^2 .

Solving this we get y = 12 y = 12 . Using the y = 12 y = 12 in equation 1 1 we get x = 20 x = 20 .

Therefore a = x 2 a = x^2 , a = 400 \boxed{a = 400}

I have made a formatting error in equation 1 it should mean a 336 = x 2 2 x y + y 2 a - 336 = x^2 - 2xy + 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

A Former Brilliant Member - 7 years, 7 months ago
Fatin Farhan
May 20, 2014

Suppose that, the 3 consecutive terms terms would be x y , x , x + y x-y ,x , x+y , ( x + y ) 2 = a + 336 (x+y)^2=a+336 , x 2 = a x^2=a , ( x + y ) 2 = a + 624 (x+y)^2=a+624 , x 2 2 x y + y 2 = 336 x^2-2xy+y^2=336 , x 2 = a , x 2 + 2 x y + y 2 = a + 624 ­ ) 2 x y + a 2 = 336 x^2=a,x^2+2xy+y^2=a+624­) \Rightarrow -2xy+a^2=-336 , 2 x y + a 2 = 624 2xy+a^2=624 , 2 a 2 = 288 a = 12 2a^2=288 \Rightarrow a=12 . x = 20 x=20 . a = 2 0 2 = 400 a=20^2=400

" 2 a 2 = 288 a = 12 2a^2=288 \Rightarrow a=12 " could be -12

Calvin Lin Staff - 7 years ago
Sukrit Kumar De
May 20, 2014

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.

"let us assume d=12; " Logically inaccurate, but a minor mistake.

Calvin Lin Staff - 7 years ago

As we have the squares of 3 consecutive numbers of an Arithmetic Progression as, ( a 336 ) , a , ( a + 624 ) (a-336),a,(a+624) Let's have ( x d ) , x , ( x + d ) (x-d),x,(x+d) as the three consecutive numbers of an arithmetic progression with d d as the common difference.Now we can write the square of the first term as, ( x d ) 2 = a 336 (x-d)^{2}=a-336 x 2 2 x d + d 2 = a 336 ( 1 ) \Rightarrow x^{2}-2xd+d^{2}=a-336-----(1) Then we can write the square of the second term as, x 2 = a ( 2 ) x^{2}=a-----(2) Then we can write the square of the third term as, ( x + d ) 2 = a + 624 (x+d)^{2}=a+624 x 2 + 2 x d + d 2 = a + 624 ( 3 ) \Rightarrow x^{2}+2xd+d^{2}=a+624-----(3) Subtracting equation (1) from equation (3),we get, 4 x d = 960 4xd=960 x d = 960 4 \Rightarrow xd=\frac{960}{4} x d = 240 ( 4 ) \Rightarrow xd=240-----(4) Subtracting equation (1) from equation (2), we get 2 x d d 2 = 336 2xd-d^{2}=336 2 × 240 336 = d 2 2 \times 240-336=d^{2} 480 336 = d 2 \Rightarrow 480-336=d^{2} d 2 = 144 \Rightarrow d^{2}=144 d = 12 \Rightarrow d=12 Substituting the value of d in equation (4),we get, x = 240 12 x=\frac{240}{12} x = 20 \Rightarrow x=20 But we know that, a = x 2 a=x^{2} a = 2 0 2 \Rightarrow a=20^{2} a = 400 \Rightarrow a=\boxed{400}

Jackal Jim
Nov 13, 2013

Let the 3 consecutive terms of the arithmetic sequence be ( b k ) , b , ( b + k ) (b-k) , b, (b+k) . Then, their squares would just be ( b 2 2 k b + k 2 ) = a 336 , b 2 = a , b 2 + 2 k b + k 2 = a + 624 (b^{2} - 2kb+k^{2}) = a - 336, b^{2} = a , b^{2}+2kb+k^{2} =a+624 .

From this, it is clear that 4 k b = 624 + 336 = 960 4kb = 624 + 336 = 960 and that 2 k b + k 2 = 624 2kb+k^{2} =624 . Then, we can know that k 2 = 624 960 2 = 144 k^{2} = 624- \frac{960}{2} = 144 , hence k = 12 k=12 . Substituting the value of k k , we can get b = 960 4 × 12 = 20 b = \frac{960}{4 \times 12} = 20 . Here, we arrive at our last step: a = b 2 = 400 a = b^{2} = \boxed {400} .

Abubakarr Yillah
Jan 13, 2014

Let the consecutive terms of the A.P be x , x + d , x + 2 d {x}, {x+d}, {x+2d}

Thus x 2 = a 336 {x^2}={a-336} , ( x + d ) 2 = a ({x+d})^2={a} and ( x + 2 d ) 2 = a + 624 ({x+2d})^2={a+624}

i.e x 2 = a 336 . . . . . ( 1 ) {x^2}={a-336}.....(1) , x 2 + 2 d x + d 2 = a . . . . . ( 2 ) {x^2}+{2dx}+{d^2}={a}.....(2) and x 2 + 4 x d + 4 d 2 = a + 624 . . . . . ( 3 ) {x^2}+{4xd}+{4d^2}={a+624}.....(3)

From.....(1) substitute (a-336) into .....(2) and .....(3)

i.e. ( a 336 ) + 2 d x + d 2 = a ({a-336})+{2dx}+{d^2}={a}

from which we get d 2 + 2 d x = 336..... ( 4 ) {d^2}+{2dx}={336}.....(4)

and ( a 336 ) + 4 x d + 4 d 2 = a + 624 ({a-336})+{4xd}+{4d^2}={a+624}

from which we get 4 d 2 + 4 d x = 960..... ( 5 ) {4d^2}+{4dx}={960}.....(5)

solving .....(4) and .....(5) simultaneously, we get d = 12 {d}={12} and x = 8 {x}={8}

but a = ( x + d ) 2 {a}=({x+d})^2

i.e. a = ( 8 + 12 ) 2 {a}=({8+12})^2

Therefore a = ( 20 ) 2 = 400 {a}=({20})^2=\boxed{400}

Tom Zhou
Dec 16, 2013

We have that a 336 , a \sqrt{a-336}, \sqrt{a} and a + 624 \sqrt{a+624} form an arithmetic progression. Therefore

2 a = a 336 + a + 624 2\sqrt{a}=\sqrt{a-336}+\sqrt{a+624}

4 a = 2 a + 288 + 2 ( a 336 ) ( a + 624 ) 4a=2a+288+2\sqrt{(a-336)(a+624)}

a 144 = ( a 336 ) ( a + 624 ) a-144=\sqrt{(a-336)(a+624)}

a 2 288 a + 20736 = a 2 + 288 a 209664 a^2-288a+20736=a^2+288a-209664

576 a = 230400 576a=230400

a = 400 a=\boxed{400}

why is root(a) multiplied by 2 in the solution

Isaac Thomas - 7 years, 5 months ago
Aritro Aich
Nov 10, 2013

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
Divyansh Joshi
May 20, 2014

(middle term-d)2=a-336 middle term squared =a (middle term +common difference )squared =a+624 a=400

Just the answer, no solution

Calvin Lin Staff - 7 years ago
Eraz Ahmed
May 20, 2014

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 .

Trial and error cannot be used here, because the numbers did not have to be integers. Because the answer was honestly found, no points either way

Calvin Lin Staff - 7 years ago
Felice Lam
Nov 16, 2013

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

Caique Ferreira
Nov 15, 2013

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

Nick Smith
Nov 13, 2013

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.

Vivien Ong
Nov 12, 2013

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 - d)^{2} , b 2 b^{2} , ( b + d ) 2 (b + d)^{2} .

After expansion from the new representation, we obtain b 2 b^{2} - 2bd + d 2 d^{2} , b 2 b^{2} , b 2 b^{2} + 2bd + d 2 d^{2} . Notice that the difference between terms are 2bd + d 2 d^{2} .

By comparing the similar progression: a - 336 , a , a + 624 and b 2 b^{2} - 2bd + d 2 d^{2} , b 2 b^{2} , b 2 b^{2} + 2bd + d 2 d^{2} , we take the differences and form two equations;

  1. - 2bd + d 2 d^{2} = -336
  2. 2bd + d 2 d^{2} = 624

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 b^{2} = a = 2 0 2 20^{2} = 400 \boxed{400}

Jian Wang
Nov 11, 2013

a = x 2 a=x^2

Therefore, a 336 , a , a + 624 = x d , x , x + d a-336,a,a+624=x-d,x,x+d

x 2 ( x d ) 2 = 336 x^2-(x-d)^2=336

x 2 x 2 d 2 + 2 d x = 336 x^2-x^2-d^2+2dx=336

d 2 + 2 d x = 336 -d^2+2dx=336

( x + d ) 2 x 2 = 624 (x+d)^2-x^2=624

x 2 + d 2 + 2 d x x 2 = 624 x^2+d^2+2dx-x^2=624

d 2 + 2 d x = 624 d^2+2dx=624

2 d 2 = 288 2d^2=288

d = 12 d=12

144 + 24 x = 624 144+24x=624

24 x = 480 24x=480

x = 20 x=20

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