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, …

Ans: Here we are given a sequence and we need to determine if the sequence qualifies to be an AP or not. After that, we need to find the common difference d and next 3 terms (after the given 4 terms).

Step 1: By observation, we have following terms in the given sequence:

First term, a1 = - \frac{1}{2}, Second term, a2 = - \frac{1}{2}, Third term, a3 = - \frac{1}{2}

Step 2: Since in an AP, the common difference d is always same between any two consecutive terms, therefore we will calculate difference between 2nd and 1st term also between 3rd term and 2nd term. Then we will check if they are equal or not.

\therefore d = a_2 - a_1 =  (- \frac{1}{2}) - (-\frac{1}{2})= 0

and d = a_3 - a_2 = (- \frac{1}{2}) - (-\frac{1}{2})= 0

Since both differences (a2 – a1) and (a3 – a2) are equal, hence the given sequence is an AP.

Step 3: Now since it is confirmed that the given sequence is an AP, we will now calculate its next 4 terms:

Here, first term a1 = - \frac{1}{2} and common difference, d = 0,

We know that the nth term of an AP is given by: an = a ( n – 1) d

Therefore, 5th term, a_5 = - \frac{1}{2} + (5 - 1) (0) = - \frac{1}{2} + 0 = - \frac{1}{2}

Similarly, 6th term, a_6 = - \frac{1}{2} + (6 - 1) (0) = - \frac{1}{2} + 0 = - \frac{1}{2}

and, 7th term, a_7 = - \frac{1}{2} + (7 - 1) (0) = - \frac{1}{2} + 0 = - \frac{1}{2}

Therefore, the given sequence is an AP, the common difference is - \frac{1}{2} and next 3 terms of this AP are: - \frac{1}{2}, - \frac{1}{2}, and - \frac{1}{2}.

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