Are -frac{1}{2}, -frac{1}{2}, -frac{1}{2}, -f | vedclass

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Question

(A) Given sequence: $-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, \ldots$To check if it is an $AP$,we calculate the difference between cons

Vedclass · Accepted Answer

Yes,$d = 0$; next three terms are $-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}$

Vedclass · Answer

(A) Given sequence: $-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, \ldots$To check if it is an $AP$,we calculate the difference between consecutive terms:$a_2 - a_1 = (-\frac{1}{2}) - (-\frac{1}{2}) = 0$$a_3 - a_2 = (-\frac{1}{2}) - (-\frac{1}{2}) = 0$$a_4 - a_3 = (-\frac{1}{2}) - (-\frac{1}{2}) = 0$Since the difference $a_{k+1} - a_k$ is constant $(0)$ for all $k$,the sequence is an $AP$ with common difference $d = 0$.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}$

Are $-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, \ldots$ in an $AP$? If they form an $AP$,find the common difference $d$ and write three more terms.

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