Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the AP:

Q : 1 Fill in the blanks in the following table, given that a is the first term, d the common difference and $\small a_n$ the $\small n$ th term of the AP:

a d n $\small a_n$

(i)

(ii)

(iii)

(iv)

(v)

7

$\small -18$

$\small ...$

$\small -18.9$

$\small 3.5$

3

$\small ...$

$\small -3$

$\small 2.5$

0

8

10

18

$\small ...$

105

$\small ...$

0

$\small -5$

$\small 3.6$

$\small ...$

Answers (1)

(i)
It is given that
$a=7, d = 3 , n = 8$
Now, we know that
$a_n = a+(n-1)d$
$a_8 = 7+(8-1)3= 7+7\times 3 = 7+21 = 28$
Therefore,
$a_8 = 28$

(ii)
It is given that
$a=-18, n = 10, a_{10} = 0$
Now, we know that
$a_n = a+(n-1)d$
$a_{10} = -18+(10-1)d$
$0 +18=9d$
$d = \frac{18}{9}=2$
(iii)
It is given that
$d=-3, n = 18, a_{18} = -5$
Now, we know that
$a_n = a+(n-1)d$
$a_{18} = a+(18-1)(-3)$
$-5=a+17\times (-3)$
$a = 51-5 = 46$
Therefore,
$a = 46$

(iv)
It is given that
$a=-18.9, d = 2.5, a_{n} = 3.6$
Now, we know that
$a_n = a+(n-1)d$
$a_{n} = -18.9+(n-1)2.5$
$3.6+18.9= 2.5n-2.5$
$n = \frac{22.5+2.5}{2.5}= \frac{25}{2.5}= 10$
Therefore,
$n = 10$

(v)
It is given that
$a=3.5, d = 0, n = 105$
Now, we know that
$a_n = a+(n-1)d$
$a_{105} = 3.5+(105-1)0$
$a_{105} = 3.5$
Therefore,
$a_{105} = 3.5$