Question

Write first four terms of the A.P. when the first term a and the common difference d are given as follows
(1) a = 10, d = 10
(2) a = − 2, d = 0
(3) a = 4, d = − 3
(4) a = − 1 d =\(\frac{1}{2}\)
(5) a = − 1.25, d = − 0.25

Answer 1

(i) a = 10, d = 10
Let the series be \(a_1 , a_2 , a_3 , a_4 , a_5\) … 
\(a_1\) = a = 10 
\(a_2 = a_1\) + d = 10 + 10 = 20 
\(a_3 = a_2\) + d = 20 + 10 = 30 
\(a_4 = a_3\) + d = 30 + 10 = 40 
\(a_5 = a_4\) + d = 40 + 10 = 50
Therefore, the series will be 10, 20, 30, 40, 50 …
⇒ The first four terms of this A.P. will be 10, 20, 30, and 40.

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(ii) a = −2, d = 0 
Let the series be \(a_1 , a_2 , a_3 , a_4\) …
\(a_1\)= a = −2
\(a_2 = a_1\) + d = − 2 + 0 = −2
\(a_3 = a_2\) + d = − 2 + 0 = −2
\(a_4 = a_3\) + d = − 2 + 0 = −2
Therefore, the series will be −2, −2, −2, −2 …
⇒ The first four terms of this A.P. will be −2, −2, −2 and −2.

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(iii) a = 4, d = −3
Let the series be \(a_1 , a_2 , a_3 , a_4\) …
\(a_1 = a\) = 4
\(a_2 = a_1\) + d = 4 − 3 = 1
\(a_3 = a_2\) + d = 1 − 3 = −2
\(a_4 = a_3\) + d = − 2 − 3 = −5
Therefore, the series will be 4, 1, −2 −5 …
⇒ The first four terms of this A.P. will be 4, 1, −2 and −5.

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(iv) a = −1, d = \(\frac{1}{2}\)
Let the series be \(a_1 , a_2 , a_3 , a_4\) … 
\(a_1 = a\) = -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 \(-1 , \frac{-1}{2},0 \text{ and }  \frac {1}{2}\)
⇒ The first four terms of this A.P. will be \(-1 , \frac{-1}{2},0\text{ and} \space \frac {1}{2}\)

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(v) a = −1.25, d = −0.25 
Let the series be \(a_1, a_2, a_3, a_4\)…
\(a_1 = a\)= −1.25
\(a_2 = a_1\) + d = − 1.25 − 0.25 = −1.50
\(a_3 = a_2\) + d = − 1.50 − 0.25 = −1.75
\(a_4 = a_3\) + d = − 1.75 − 0.25 = −2.00
Clearly, the series will be 1.25, −1.50, −1.75, −2.00 …….. 
⇒ The first four terms of this A.P. will be −1.25, −1.50, −1.75 and −2.00.