If the sum of four numbers in A.P is 24 and the sum of their squares 164, then those numbers are 

Correct option is (A) 3, 5, 7, 9

Let a-3d, a-d, a+d, a+3d are required 4 numbers in A.P.

Given that sum of these four numbers is 24.

i.e., (a-3d) + (a-d) + (a+d) + (a+3d) = 24

\(\Rightarrow4a=24\)

\(\Rightarrow a=\frac{24}4=6\)

Also given that the sum of their squares is 164.

\(i.e.,(a-3d)^2+(a-d)^2\) \(+(a+d)^2+(a+3d)^2=164\)

\(\Rightarrow(a-3d)^2+(a+3d)^2\) \(+(a-d)^2+(a+d)^2=164\)

\(\Rightarrow2(a^2+(3d)^2)+2(a^2+d^2)=164\)     \((\because(a-b)^2+(a+b)^2=2(a^2+b^2))\)

\(\Rightarrow(a^2+9d^2)+(a^2+d^2)=\frac{164}2=82\)

\(\Rightarrow2a^2+10d^2=82\)

\(\Rightarrow10d^2=82-2a^2\)

\(\Rightarrow10d^2=82-2\times6^2\)       \((\because a=6)\)

\(=82-72=10\)

\(\Rightarrow d^2=\frac{10}{10}=1\)

\(\therefore d=1\)

Now, \(a-3d=6-3=3,\)

\(a-d=6-1=5,\)

\(a+d=6+1=7\)

and \(a+3d=6+3=9\)

Hence, required numbers are 3, 5, 7, 9.

Correct option is A) 3, 5, 7, 9