Question.
For the following APs, write the first term and the common difference :
(i) $3,1,-1,-3, \ldots$
(ii) $-5,-1,3,7, \ldots \ldots$
(iii) $\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots$
(iv) $0.6,1.7,2.8,3.9, \ldots$
For the following APs, write the first term and the common difference :
(i) $3,1,-1,-3, \ldots$
(ii) $-5,-1,3,7, \ldots \ldots$
(iii) $\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots$
(iv) $0.6,1.7,2.8,3.9, \ldots$
Solution:
(i) $\mathrm{a}=3, \mathrm{~d}=\mathrm{t}_{2}-\mathrm{t}_{1}=1-3=-2$,
i.e., $d=-2$
(ii) $a=-5, d=4$
(iii) $a=\frac{1}{3}$
$\mathrm{d}=\mathrm{t}_{2}-\mathrm{t}_{1}=\frac{5}{3}-\frac{1}{3}=\frac{4}{3}$
(iv) $0.6,1.7,2.8,3.9 \ldots$
$a=0.6$
$\mathrm{d}=\mathrm{t}_{2}-\mathrm{t}_{1}$
$=1.7-0.6$
$=1.1$
(i) $\mathrm{a}=3, \mathrm{~d}=\mathrm{t}_{2}-\mathrm{t}_{1}=1-3=-2$,
i.e., $d=-2$
(ii) $a=-5, d=4$
(iii) $a=\frac{1}{3}$
$\mathrm{d}=\mathrm{t}_{2}-\mathrm{t}_{1}=\frac{5}{3}-\frac{1}{3}=\frac{4}{3}$
(iv) $0.6,1.7,2.8,3.9 \ldots$
$a=0.6$
$\mathrm{d}=\mathrm{t}_{2}-\mathrm{t}_{1}$
$=1.7-0.6$
$=1.1$
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