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If the sum of four numbers in A.P is 24 and the sum of their squares 164, then those numbers are A) 3, 5, 7, 9 B) 3, 4, 5, 6 C) 1, 4, 7,11 D) 2, 4, 6,12 |
Answer» 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 |
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