Question

If $a_1,a_2,a_3,\cdots$ are in A.P. and
$a_1^2-a_2^2+a_3^2-a_4^2+\cdots +a_{2k-1}^2-a_{2k}^2$
$=M(a_1^2-a_{2k}^2)$. Then M =

• A. $\frac{k-1}{k+1}$
• B. $\frac{k}{2k-1}$
• C. $\frac{k+1}{2k+1}$
• D. None

Answer 1

The correct option is B $\frac{k}{2k-1}$

$a_2-a_1=a_3-a_2=\cdots a_{2k}-a_{2k-1}=dHence,a_1^2-a_2^2=(a_1-a_2)(a_1+a_2)=-d(a_1+a_2)a_3^2-a_4^2=(a_3-a_4)(a_3+a_4)=-d(a_3+a_4)...............................................................a_{2k-1}^2-a_{2k}^2=\left(a_{2k-1}{a}_{2}k\right)(a_{2}k-1+a_{2k})=-d(a_{2k-1}+a_{2k})$
Adding, we get
$a_1^2-a_2^2+a_3^2-a_4^2+\cdots a_{2k-1}^2-a_{2k}^2=-d(a_1+a_2+a_3+a_4+\cdots a_{2k-1}+a_{2k})=-d.\frac{2k}{2}(a_1+a_{2k})=-dk(a_1+a_{2k})$
But $a_{2k}=a_1+(2k-1)d\Rightarrow -d=\frac{a_1-a_{2k}}{2k-1}$
$\therefore$ The required sum= $\frac{k}{2k-1}(a_1^2-a_{2k}^2)\Rightarrow M=\frac{k}{2k-1}$