Find four numbers in A P whose sum is 20 and sum of square is 120 - Maths - Arithmetic Progressions

Question

Find four numbers in A P whose sum is 20 and sum of square is
120

Ajanta Trivedi · Accepted Answer

let the numbers are (a-3d),(a-d),(a+d) and (a+3d)

sum of the numbers is 20.

therefore

$a - 3 d + (a - d) + (a + d) + (a + 3 d) = 20 4 a = 20 \Rightarrow a = \frac{20}{4} = 5$

therefore the numbers are (5-3d),(5-d),(5+d) and (5+3d)

since sum of the square is 120.

$(5 - 3 d)^{2} + (5 - d)^{2} + (5 + d)^{2} + (5 + 3 d)^{2} = 120 [(5 - 3 d)^{2} + (5 + 3 d)^{2}] + [(5 - d)^{2} + (5 + d)^{2}] = 120$ ............(1)

since $(a - b)^{2} + (a + b)^{2} = 2 (a^{2} + b^{2})$

the equation (1) becomes:

$2 (25 + 9 d^{2}) + 2 (25 + d^{2}) = 120 50 + 18 d^{2} + 50 + 2 d^{2} = 120 20 d^{2} = 120 - 100 = 20 d^{2} = \frac{20}{20} = 1 d = \pm 1$

therefore the numbers are $(5 - 3), (5 - 1), (5 + 1) a n d (5 + 3)$

i.e. numbers are 2,4,6,8. or 8,6,4,2

Find four numbers in A.P whose sum is 20 and sum of square is 120..