Match the questions given under Column I with their appropriate a… - Vidyadip

Question

Match the questions given under Column I with their appropriate answers given under the Column $II.$

	$\text{Column I} $		$\text{Column II} $
$(a)$	$4,1,\frac{1}{4},\frac{1}{16}$	$(i)$	$A.P.$
$(b)$	$2, 3, 5, 7$	$(ii)$	$Squence$
$(c)$	$13, 8, 3, -2, -7$	$(iii)$	$G.P.$

✓

Answer

	$\text{Column I}$		$\text{Column II}$
$(a)$	$4,1,\frac{1}{4},\frac{1}{16}$	$(i)$	$G.P.$
$(b)$	$2, 3, 5, 7$	$(ii)$	$Squence$
$(c)$	$13, 8, 3, -2, -7$	$(iii)$	$A.P.$

$4,1,\frac{1}{4},\frac{1}{16}$
Here, $\frac{\text{a}_2}{\text{a}_1}=\frac{1}{4},\frac{\text{a}_3}{\text{a}_2}=\frac{\frac{1}{4}}{1}=\frac{1}{4}$ and $\frac{\text{a}_4}{\text{a}_3}=\frac{\text{a}_4}{\text{a}_3}=\frac{\frac{1}{16}}{\frac{1}{4}}=\frac{1}{4}$
Hence, it is $G.P.$
$2, 3, 5, 7$
Here$, a2 - a1 = 3 - 2 = 1 a3 - a2 = 5 - 3 = 2 \therefore\ \text{a}_2-\text{a}_1\neq\text{a}_3-\text{a}_2$ Hence$,$ it is not $A.P. \frac{\text{a}_2}{\text{a}_1}=\frac{3}{2},\frac{\text{a}_3}{\text{a}_2}==\frac{5}{3}$
So $,\frac{3}{2}\neq\frac{5}{3}$
So$,$ it is not $G.P.$ Hence, it is squence.
$13, 8, 3, -2, -7$
Here$, a3 - a1 = 8 - 13 = -5 a3 - a2 = 3 - 8 = -5 So, a2 - a1 = a3 - a2 = -5$
So $,$ it is an $A.P.$

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