Question

Is the following sequence an $AP$? If true, find the common difference $d$ and write three more terms.
$-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},.$

Answer 1

$-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}$

$\Rightarrow$ Difference between second and first term $=-\frac{1}{2}-(-\frac{1}{2})==-\frac{1}{2}+\frac{1}{2}=0$

$\Rightarrow$ Difference between third and second term $=-\frac{1}{2}-(-\frac{1}{2})==-\frac{1}{2}+\frac{1}{2}=0$

$\Rightarrow$ Difference between fourth and third term $=-\frac{1}{2}-(-\frac{1}{2})==-\frac{1}{2}+\frac{1}{2}=0$

$\Rightarrow$ Since, difference is same, it is an $A.P$

$\Rightarrow$ Common difference $=d=0$

$\Rightarrow$ We have to find next three terms. We are given $4$ terms.

$\Rightarrow$ We have to find $5^{th},6^{th}$ and $7^{th}$ terms.

$\Rightarrow$ $5^{th}$ term $=4^{th}term+d=-\frac{1}{2}+0=-\frac{1}{2}$

$\Rightarrow$ $6^{th}$ term $=5^{th}term+d=-\frac{1}{2}+0=-\frac{1}{2}$

$\Rightarrow$ $7^{th}$ term $=6^{th}term+d=-\frac{1}{2}+0=-\frac{1}{2}$

Hence $5^{th},6^{th}$ and $7^{th}$ terms are $-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}$