Question - 2 (iv)
Write first four terms of the A.P. when the first term "a" and the common difference "d" are given as follows.
(iv) $a = - 1, d = \frac{1}{2}$

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Solution

(iv)
$a = - 1, d = \frac{1}{2} L e t t h e s e r i e s b e a_{1}, a_{2}, a_{3}, a_{4} \dots a_{1} = a = - 1 a_{2} = a_{1} + d = - 1 + \frac{1}{2} = - \frac{1}{2} a_{3} = a_{2} + d = - \frac{1}{2} + \frac{1}{2} = 0 a_{4} = a_{3} + d = 0 + \frac{1}{2} = \frac{1}{2}$
Clearly, the series will be $- 1, - \frac{1}{2}, 0, \frac{1}{2}$
First four terms of this A.P. will be $- 1, - \frac{1}{2}, 0 a n d \frac{1}{2}$

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Question - 2 (iv) Write first four terms of the A.P. when the first term "a" and the common difference "d" are given as follows. (iv) a=−1, d=12

(iv) a=−1,d=12Let the series be a1,a2,a3,a4…a1=a=−1a2=a1+d=−1+12=−12a3=a2+d=−12+12=0a4=a3+d=0+12=12 Clearly, the series will be −1,−12,0,12 First four terms of this A.P. will be −1,−12,0 and 12

Question - 2 (iv)
Write first four terms of the A.P. when the first term "a" and the common difference "d" are given as follows.
(iv) $a = - 1, d = \frac{1}{2}$