Question

In the following table, given that $a$ is the first term, $d$ the common difference and $a_n$ the $nth$ term of the AP.

	$a$	$d$	$n$	$a_n$
(i)	7	3	8	.......
(ii)	-18	.......	10	0
(iii)	.......	-3	18	-5
(iv)	-18.9	2.5	.......	3.6
(v)	3.5	0	105	.......

• A. $\left(i\right)a_n=13\left(ii\right)d=1\left(iii\right)a=46\left(iv\right)n=9\left(v\right)a_n=3.5$
• B. $\left(i\right)a_n=36\left(ii\right)d=7\left(iii\right)a=46\left(iv\right)n=13\left(v\right)a_n=3.5$
• C. $\left(i\right)a_n=18\left(ii\right)d=3\left(iii\right)a=46\left(iv\right)n=4\left(v\right)a_n=3.5$
• D. $\left(i\right)a_n=28\left(ii\right)d=2\left(iii\right)a=46\left(iv\right)n=10\left(v\right)a_n=3.5$

Answer 1

Answer: (D)

$\left(i\right)a_n=28\left(ii\right)d=2\left(iii\right)a=46\left(iv\right)n=10\left(v\right)a_n=3.5$

Here we use the formula $t_n=a+(n-1)d$

(i). $a_8=a+(8-1)d=7+7\ast 3=28$
(ii). $a=-18,d=?$
$t_n=1+(n-1)d$
$0=-18+(10-1)d$
$d=2$
(iii). $-5=a+(18-1)(-3)$
Solve this to get a=46.
(iv). $3.6=-18.9+(n-1)2.5$
n= 10.
(v). $t_n=3.5+(105-1)0=3.5$