Question 4 (iv)
Which of the following are APs? If they form an A.P. find the common difference d and write three more terms.
(iv) -10, - 6, - 2, 2 …

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Solution

$- 10, - 6, - 2, 2 \dots \dots \dots$
Here,
$a_{2} - a_{1} = (- 6) - (- 10) = 4 a_{3} - a_{2} = (- 2) - (- 6) = 4 a_{4} - a_{3} = (2) - (- 2) = 4$
$\Rightarrow a_{n + 1} - a_{n}$ is same everytime.
Therefore, d=4 and the given numbers are in A.P.
Next three terms are:
$a_{5} = 2 + 4 = 6 a_{6} = 6 + 4 = 10 a_{7} = 10 + 4 = 14$

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Question 4 (iv) Which of the following are APs? If they form an A.P. find the common difference d and write three more terms. (iv) -10, - 6, - 2, 2 …

−10,−6,−2,2……… Here, a2−a1=(−6)−(−10)=4a3−a2=(−2)−(−6)=4a4−a3=(2)−(−2)=4 ⇒an+1−an is same everytime. Therefore, d=4 and the given numbers are in A.P. Next three terms are: a5=2+4=6a6=6+4=10a7=10+4=14

Question 4 (iv)
Which of the following are APs? If they form an A.P. find the common difference d and write three more terms.
(iv) -10, - 6, - 2, 2 …