Question

The sum of four consecutive numbers in an A.P. with common difference $d>0$ is 20. If the sum of their squares is 120, then the middle terms are __ and __.

• A. 2,4
• B. 4,6
• C. 6,8
• D. 8,10

Answer 1

The correct option is

A

4,6

Let the numbers be $a-3d,a-d,a+d$ and $a+3d.$ These numbers form an AP with first term $a-3d$ and common difference $d>0.$
Also, given that $(a-3d)+(a-d)+(a+d)+(a+3d)=20.$
$\Rightarrow 4a=20$
$\Rightarrow a=5$
Also,

${(a-3d)}^2+{(a-d)}^2+{(a+d)}^2+{(a+3d)}^2=120$

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

$\Rightarrow 4a^2+20d^2=120$
$\Rightarrow 4(5{)}^2+20d^2=120$

$\Rightarrow 4\left(25\right)+20d^2=120$

$\Rightarrow d^2=1$
$\Rightarrow d=\pm 1$
SIince $d>0,$ we must have $d=1.$

1st number = $a-3d=5-3\times 1=2$

2nd number = $a-d=5-1=4$

3rd number = $a+d=5+1=6$

4th number = $a+3d=5+3\times 1=8$
Hence, the numbers are 2, 4, 6, and 8.