Question

Find the four number in

a.p. whose sum is 20 and the sum of there squares is 120

Answer 1

let no. be (a-d),(a-2 d),(a+d),(a+2 d)

so,as per first condition ,

4a=20

a=5

solve for second condition

Answer 2

AnswEr:

Let the numbers be (a-3d) , (a-d), (a+d), (a+3d). Then,

Sum = 20

$\tt \implies \: (a - 3d) + (a - d) + (a + d) + (a + 3d) = 20 \\ \\ \tt \implies \: 4a = 20 \\ \\ \tt \implies a = \cancel\frac {20}{4} = 5 \\ \\ \implies \tt \: a = 5$

and, Sum of the squares = 120

$\implies \tt \: {(a - 3d)}^{2} + {(a - d)}^{2} + {(a + d)}^{2} + {(a + 3d)}^{2} = 120 \\ \\ \tt \implies \: 4 {a}^{2} + 20 {d}^{2} = 120 \\ \\ \implies \tt {a}^{2} + 5 {d}^{2} = 30 \\ \\ \tt \implies \: 25 + 5 {d}^{2} = 30 \\ \\ \tt \implies \: 5 {d}^{2} = 5 \\ \\ \implies \tt \: d = \underline + 1$

If d = 1, and a = 5, then the numbers are 2,4,6,8. If d = -1, and a = 5, then the numbers are 8,6,4,2.

• Thus, the numbers are 2,4,6,8 or 8,6,4,2.