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},...

Answers (1)

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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