Question

Question 1
Match the AP’s given in column A with suitable common differences given in column B.
$\begin{array}{cccc}& ColumnA& & ColumnB\\ \left(A_1\right)& 2,-2,-6,-10,....& \left(B_1\right)& \frac{2}{3}\\ \left(A_2\right)& a=-18,n=10,a_n=0& \left(B_2\right)& -5\\ \left(A_3\right)& a=0,a_{10}=6& \left(B_3\right)& 4\\ \left(A_4\right)& a_2=13,a_4=3& \left(B_4\right)& -4\\ & & \left(B_5\right)& 2\\ & & \left(B_6\right)& \frac{1}{2}\\ & & \left(B_7\right)& 5\end{array}$

Answer 1

$\left(A_1\right)-\left(B_4\right)$
2, - 2, - 6, - 10, ...
Here, common difference, d = - 2 - 2 = - 4

$\left(A_2\right)-\left(B_5\right)$
$\because a_n=a+(n-1)d$
$\Rightarrow$ 0 = -18 + (10 - 1)d
18 = 9d
$\therefore$ Common difference, d = 2

$\left(A_3\right)-\left(B_1\right)$
$\because$ $a_{10}=6$
$\Rightarrow$ a + (10 - 1)d = 6
$\Rightarrow$ 0 + 9d = 6 [$\because$ a = 0 (given)]
$\Rightarrow 9d=6\Rightarrow d=\frac{2}{3}$

$\left(A_4\right)-\left(B_2\right)$
because~a_2=13\)
$\Rightarrow a+(2-1)d=13[\because a_n=a+(n-1\left)d\right]$
$\Rightarrow a+d=13\cdots \left(i\right)$
$and{a}_4=3\Rightarrow a+(4-1)d=3$
$\therefore a+3d=3\cdots \left(ii\right)$
On subtracting eq.(i) from eq.(ii), we get;
2d = -10
$\Rightarrow$ d = -5

$\therefore \left(A_1\right)\to \left(B_4\right),\left(A_2\right)\to \left(B_5\right),\left(A_3\right)\to \left(B_1\right)and\left(A_4\right)\to \left(B_2\right)$