Question

Find 4 numbers in ap whose sum is 20 and the sum of whose square is 120

Answer 1

Let the four numbers in A.P be a-3d, a-d,a+d,a+3d.    ---- (1)

Given that Sum of the terms = 20.

= (a-3d) + (a-d) + (a+d) + (a+3d) = 20

4a = 20

  a = 5.    ---- (2)


Given that sum of squares of the term = 120.

= (a-3d)^2 + (a-d)^2 + (a+d)^2 + (a+3d)^2 = 120

= (a^2 + 9d^2 - 6ad) + (a^2+d^2-2ab) + (a^2+d^2+2ad) + (a^2+9d^2+6ad) = 120

= 4a^2 + 20d^2 = 120

Substitute a = 5 from (2) .

4(5)^2 + 20d^2 = 120

100 + 20d^2 = 120

20d^2 = 20

d = +1 (or) - 1.

Since AP cannot be negative.

Substitute a = 5 and d = 1 in (1), we get

a - 3d, a-d, a+d, a+3d = 2,4,6,8.


Hope this helps!

Answer 2

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student-name Anushka Saini asked in Math

FIND FOUR NUMBERS IN AP WHOSE SUM IS 20 &SUM OF WHOSE SQUARES IS 120

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student-name Sachin Bagga answered this

26 helpful votes in Math, Class XII-Science

let four numbers in a.p. be:  

a-3d,a-d,a+d,a+3d

their sum=(a-3d)+(a-d)+

(a+d)+(a+3d)=20

=> 4a=20  =>a=5

also sum of squares=120

i.e.  (a-3d)2+(a-d)2

+(a+d)2+(a+3d)2=120

 {where  2 means square }

=>a2-6ad+9d2-2ad +d2+

a2+2ad+d2+a2+6ad+9d2=120

=>4a2=20d2=120

=>4(25)=20d2=120

=>20d2+20  =>d2=1

=>d=+_1

hence the numbers are

5-3,5-1,5-1,5-3

or  5-3,5-1,5-1,5-3

i.e.  2,4,6,8  or8,6,4,2