Question

Verify that following is an AP, and then write its next three terms.

$0,\frac{1}{4},\frac{1}{2},\frac{3}{4},....$

Answer 1

We have,
$\frac{1}{4}-0=\frac{1}{4};$

$\frac{1}{2}-\frac{1}{4}=\frac{1}{4};$

This shows that the difference of a term and the preceding term is always the same. Hence, the given sequence form an AP.

nth term $t_n=a+(n-1)d$

$a=$ First term

$d=$ Common difference

$n=$ number of terms

$t_n=n^{th}$ term

Here first term $a=0$ and common difference $d=\frac{1}{4}$
Therefore,
$a_5=a+4d=0+4\times \frac{1}{4}$
$\Rightarrow a_5=0+1=1$

$a_6=a+5d=0+5\times \frac{1}{4}$
$\Rightarrow a_5=\frac{5}{4}$

$a_7=a+6d=0+6\times \frac{1}{4}$
$\Rightarrow a_7=\frac{3}{2}$

Hence, the next three terms are $1,\frac{5}{4},\frac{3}{2}$