Question

P,q,r,s,and t are 5 consecutive positive integers what are they?

P²+Q²+R²=S²+T²​

Answer 1

HEY MATE YOUR ANSWER IS HERE...

LET 5 CONSECUTIVE TERMS BE :-

$x = p\\ x + 1 = q\\ x + 2 = \: r\\ x + 3 = s \\ x + 4 = t$

NOW ACCORDING TO THE QUESTION :-

P² + Q² + R² = S² + T²

HENCE ,

${x}^{2} + {(x + 1)}^{2} + (x + 2) ^{2} </strong><strong>\</strong><strong>\</strong><strong>\</strong><strong> </strong><strong>= {(x + 3) + (x + 4)}^{2}$

HENCE ,

${x}^{2} + {x}^{2} + 1 + 2x + {x}^{2} + 4 + 4x \\ = {x}^{2} + 9 + 6x + {x}^{2} + 16 + 8x$

NOW BY FURTHER CALCULATION

$3 {x}^{2} + 5 + 6x = 2 {x}^{2} + 25 + 14x$

NOW BY CANCELLING OUT TERMS

${x}^{2} - 8x - 20 = 0$

NOW BY MIDDLE TERM SPLIT

${x}^{2} - 10x + 2x - 20$

$x(x - 10) + 2(x - 10) = 0$

HENCE ,

$(x - 10)(x + 2) = 0$

NOW VALUE OF X IS

$x = 10 \: and \: \: x = - 2$

HENCE

IF X = 10

THEN YOUR VALUES ARE

10 , 11 ,12 , 13 , 14

10² + 11² + 12² = 13² + 14²

100 + 121 + 144 = 169 + 196

365 = 365

NOW LET X = -2

THEN YOUR VALUES ARE

-2 , -1 ,0 ,1 , 2

(-2)² +( -1)² + 0² = 1² + 2²

4 + 1 + 0 = 4 + 1

5 = 5

SO UR VALUES ARE

10 , 11 ,12 , 13 , 14

AND

-2 , -1 ,0 ,1 , 2

THANKS FOR YOUR QUESTION HOPE THIS HELPS

★ KEEP SMILING ☺️✌️ ★

Answer 2

Given:

P, Q, R, S, T are 5 consecutive integers

To Find:

the values of P²+ Q²+ R²= S²+ T²

Solution:

    P²+ Q²+ R² = S²+ T²

  ⇒ x²+(x+1)²+(x+2)²= (x+3)²+ (x+4)²

  ⇒ 3x²+ 6x+ 5 = 2x²+13x+25

  ⇒ x²+8x = 20

  ⇒ x²+ 8x+ 20 = 0

Now, factorize by middle term splitting

    ⇒ x²+ 10x- 2x- 20 = 0

    ⇒ x(x+10)- 2(x+10) = 0

    ⇒ (x-2)(x+10) = 0

    Now, x can either be 2 or -10

as we know that it is a positive integer, then

       x= 2

So, the numbers are 2, 3, 4, 5, 6 for P, Q, R, S T respectively.