Fill in the blanks in the following table,

Question.

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

Solution:

(i) $\mathrm{a}=7, \mathrm{~d}=3, \mathrm{n}=8$

$a_{8}=a+7 d=7+7 \times 3=28$

Hence, $a_{8}=28$.

(ii) $\mathrm{a}=-18, \mathrm{n}=10, \mathrm{a}_{\mathrm{n}}=0, \mathrm{~d}=?$

$a_{n}=a+(n-1) d$

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

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

Hence, $d=2$

(iii) $d=-3, n=18, a_{n}=-5$

$a_{n}=a+(n-1) d$

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

$-5=a+(17)(-3)$

$-5=a-51$

$a=51-5=46$

Hence, $a=46$

(iv) $a=-18.9, d=2.5$

$\mathrm{t}_{\mathrm{n}}=3.6$

$\Rightarrow a+(n-1) d=3.6$

$\Rightarrow-18.9+(n-1) \times(2.5)=3.6$

$\Rightarrow(\mathrm{n}-1) \times(2.5)=3.6+18.9=22.5$

$\Rightarrow \mathrm{n}-1=\frac{22.5}{2.5}=\frac{225}{25}=9$

$\Rightarrow \mathrm{n}=10$

(v) $\mathrm{a}=3.5, \mathrm{~d}=0, \mathrm{n}=105$

Then $\mathrm{a}_{105}=\mathrm{a}+104 \mathrm{~d}=3.5+0=3.5$