Q

Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. (viii) -1/2, -1/2, -1/2, 1/2, ...

Q : 4        Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.

(viii) $\small -\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},...$

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Given series is
$\small -\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},...$
Now,
the first term to this series is = $-\frac{1}{2}$
Now,
$a_1 = -\frac{1}{2} \ \ and \ \ a_2 = -\frac{1}{2} \ \ and \ \ a_3 = -\frac{1}{2} \ \ and \ \ a_4 = -\frac{1}{2}$
$a_2-a_1 = -\frac{1}{2}-\left ( -\frac{1}{2} \right ) = -\frac{1}{2}+\frac{1}{2}=0$
$a_3-a_2 = -\frac{1}{2}-\left ( -\frac{1}{2} \right ) = -\frac{1}{2}+\frac{1}{2}=0$
$a_4-a_3 = -\frac{1}{2}-\left ( -\frac{1}{2} \right ) = -\frac{1}{2}+\frac{1}{2}=0$
We can clearly see that the difference between terms are equal  and equal to 0
Hence, given series is in AP
Now, the next three terms are
$a_5=a_4+d = -\frac{1}{2}+0=-\frac{1}{2}$
$a_6=a_5+d = -\frac{1}{2}+0=-\frac{1}{2}$
$a_7=a_6+d = -\frac{1}{2}+0=-\frac{1}{2}$
Therefore, the next three terms of given series are  $-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}$

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