Question

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

Column A	Column B
(A1) $2,–2,–6,–10,\dots$	(B1) $\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$
(A2) $\mathrm{a}=–18,\mathrm{n}=10,\mathrm{an}=0$	(B2) $–5$
(A3) $\mathrm{a}=0,\mathrm{a}10=6$	(B3) $4$
(A4) $\mathrm{a}2=13,\mathrm{a}4=3$	(B4) $–4$
(B5) $2$
(B6) $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$
(B7)$5$

Answer 1

The explanation for $\left(a\right)$ :

Compute the common difference:

$2,–2,–6,–10,\dots$

Common Difference $={\mathrm{a}}_2-{\mathrm{a}}_1$

$=-2-2=-4$

Hence $\left(a\right)\to \left(iv\right)$

Explanation for $\left(b\right)$:

Compute the common difference:

$\mathrm{a}=–18,\mathrm{n}=10,\mathrm{an}=0$

The formula for the nth term of AP

$$\begin{gathered} \Rightarrow {\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1)\mathrm{d} \\ \Rightarrow 0=-18+(10-1)\mathrm{d} \\ \Rightarrow 18=9\mathrm{d} \\ \Rightarrow \frac{18}{9}=\mathrm{d} \\ \Rightarrow 2=\mathrm{d} \end{gathered}$$

Hence $\left(b\right)\to \left(iii\right)$

The explanation for $\left(c\right)$:

Compute the common difference:

$\mathrm{a}=0,{\mathrm{a}}_{10}=6$

$$\begin{gathered} {\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1)\mathrm{d} \\ \Rightarrow {\mathrm{a}}_{10}=\mathrm{a}+(10-1)\mathrm{d} \\ \Rightarrow 6=0+9\mathrm{d} \\ \Rightarrow \frac{6}{9}=\mathrm{d} \\ \Rightarrow \frac{2}{3}=\mathrm{d} \end{gathered}$$

Hence $\left(c\right)\to \left(i\right)$

The explanation for $\left(d\right)$:

Compute the common difference:

${\mathrm{a}}_2=13,{\mathrm{a}}_4=3$

$$\begin{gathered} {\mathrm{a}}_2=\mathrm{a}+\mathrm{d}{\mathrm{a}}_4=\mathrm{a}+3\mathrm{d} \\ \mathrm{a}={\mathrm{a}}_2-\mathrm{d}-\left(1\right)\mathrm{a}={\mathrm{a}}_4-3\mathrm{d}-\left(2\right) \\ \mathrm{From}\left(1\right)\mathrm{and}\left(2\right) \\ \Rightarrow {\mathrm{a}}_2-\mathrm{d}={\mathrm{a}}_4-3\mathrm{d} \\ \Rightarrow 13-\mathrm{d}=3-3\mathrm{d} \\ \Rightarrow 13-3=\mathrm{d}-3\mathrm{d} \\ \Rightarrow 10=-2\mathrm{d} \\ \Rightarrow -5=\mathrm{d} \end{gathered}$$

Hence $\left(d\right)\to \left(ii\right)$

Hence $\left(a\right)\to \left(iv\right),\left(b\right)\to \left(iii\right),\left(c\right)\to \left(i\right),\left(d\right)\to \left(ii\right)$.