Match the APs given in column A with suitable common differences … - Vidyadip

Question

Match the APs given in column A with suitable common differences given in column B.

Column A		Column B
$A_1$	$2, -2, -6, -10$	$B_1$	$\frac{2}{3}$
$A_2$	$a = -18, n = 10, a_n = 0$	$B_2$	-5
$A_3$	$a = 0, a_{10} = 6$	$B_3$	4
$A_4$	$a_2= 13, a_4 = 3$	$B_4$	-4
		$B_5$	2
		$B_6$	$\frac{1}{2}$
		$B_7$	5

✓

Answer

Column A		Column B
$A_1$	$2, -2, -6, -10$	$B_4$	$-4$
$A_2$	$a = -18, n = 10, a_n = 0$	$B_5$	$2$
$A_3$	$a = 0, a_{10} = 6$	$B_1$	$\frac{2}{3}$
$A_4$	$a_2= 13, a_4 = 3$	$B_2$	$-5$

$A_1, 2, -2, -6, -10, ......$
Here, Common difference, d = -2 - 2 = -4
$A_2$ $\because$ $a_1 = a + (n - 1)d$
$\Rightarrow 0 = -18 + (10 - 1)d$
$\Rightarrow 18 = 9d$
$\therefore$ Common difference, d = 2
$A_3$ $\because$ $a_{10} = 6$
$\Rightarrow a + (10 - 1)d = 6$
$\Rightarrow 0 + 9d = 6$ [$\because$ a = 0 (given)]
$\Rightarrow 9d = 6$
$\Rightarrow\text{d}=\frac{2}{3}$
$A_4$ $\because$ $a_2 = 13$
[$\because$ $a_n = a + (n - 1)d]$
$\Rightarrow a + (2 - 1)d = 13$
$\Rightarrow a + d = 13 ......(i)$
and $a_4= 3$
$\Rightarrow a + (4 - 1)d = 3$
$\Rightarrow a + 3d = 3 ........(ii)$
on subtracting Eq. (i) from Eq. (ii), we get-
$\Rightarrow 2d = -10$
$\Rightarrow d = -5$

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