Question

Sr. no.	$a$	$d$	$n$	$a_n$
(i)	7	3	8	----
(ii)	-18	----	10	0
(iii)	----	-3	18	-5
(iv)	-18.9	2.5	----	3.6
(v)	3.5	0	105	----

Fill in the blanks in the following table, given that $a$ is the first term, $d$ the common difference and $a_n$ is the $n^{th}$ term of the AP:

Answer 1

(i) $a=7,d=3,n=8,a_n=?$

We know that,

For an A.P. $a_n=a+(n-1)d$

$=7+(8-1)3$

$=7+\left(7\right)3$

$=7+21=28$

Hence, $a_n=28$

(ii) Given that

$a=-18,n=10,a_n=0,d=?$

We know that,

$a_n=a+(n-1)d$

$0=-18+(10-1)d$

$18=9d$

$d=\frac{18}{9}=2$

Hence, common difference, $d=2$

(iii) Given that

$d=-3,n=18,an=-5$

We know that,

$a_n=a+(n-1)d$

$-5=a+(18-1)(-3)$

$-5=a+\left(17\right)(-3)$

$-5=a-51$

$a=51-5=46$

Hence, $a=46$

(iv) $a=-18.9,d=2.5,a_n=3.6,n=?$

We know that,

$a_n=a+(n-1)d$

$3.6=-18.9+(n-1)2.5$

$3.6+18.9=(n-1)2.5$

$22.5=(n-1)2.5$

$(n-1)=\frac{22.5}{2.5}=9$

$n-1=9$

$n=10$

Hence, $n=10$

(v) $a=3.5,d=0,n=105,a_n=?$

We know that,

$a_n=a+(n-1)d$

$a_n=3.5+(105-1)0$

$a_n=3.5+104\times 0$

$a_n=3.5$