Ahmad's sequence of squares

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

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, $d = x_2-x_1 = x_3-x_2$

$x_2-x_1 = x_3-x_2$

$x_2+x_2 = x_3+x_1$

${(2x_2)}^2 = {(x_3+x_1)}^2$

$4{x_2}^2 = {x_3}^2+2x_1x_3+{x_1}^2$

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

thus, $x_1 = \sqrt{a-336}; x_3 = \sqrt{a+624}$

$4{x_2}^2 = {x_3}^2+2x_1x_3+{x_1}^2$

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

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

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

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

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

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

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

${576a} = {230400}$

${a} = {400}$ 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 $a - 336$ , $a$ and $a + 624$ 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 $3$ $a + 624 = x^2 + 2xy + y^2$ .

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

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

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

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

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

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

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, $(a-336),a,(a+624)$ 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, $(x-d)^{2}=a-336$ $\Rightarrow x^{2}-2xd+d^{2}=a-336-----(1)$ 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, $(x+d)^{2}=a+624$ $\Rightarrow x^{2}+2xd+d^{2}=a+624-----(3)$ Subtracting equation (1) from equation (3),we get, $4xd=960$ $\Rightarrow xd=\frac{960}{4}$ $\Rightarrow xd=240-----(4)$ Subtracting equation (1) from equation (2), we get $2xd-d^{2}=336$ $2 \times 240-336=d^{2}$ $\Rightarrow 480-336=d^{2}$ $\Rightarrow d^{2}=144$ $\Rightarrow d=12$ Substituting the value of d in equation (4),we get, $x=\frac{240}{12}$ $\Rightarrow x=20$ But we know that, $a=x^{2}$ $\Rightarrow a=20^{2}$ $\Rightarrow a=\boxed{400}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$576a=230400$

$a=\boxed{400}$

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;

1. - 2bd + $d^{2}$ = -336
2. 2bd + $d^{2}$ = 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, $b^{2}$ = a = $20^{2}$ = $\boxed{400}$

$a=x^2$

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

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

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

$-d^2+2dx=336$

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

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

$d^2+2dx=624$

$2d^2=288$

$d=12$

$144+24x=624$

$24x=480$

$x=20$