Question

Write first four terms of the AP, when the first term a and the common difference d are given as follows:
(i). \[{\rm{a}} = {\rm{1}}0\], \(d = 10\)
(ii). \[{\rm{a}} = - 2\], \(d = 0\)
(iii). \[{\rm{a}} = 4\], \(d = - 3\)
(iv). \[{\rm{a}} = - 1\], \(d = \dfrac{1}{2}\)
(v). \[{\rm{a}} = - 1.25\], \(d = - 0.25\)

Answer 1

Hint: If the first term and the common difference of an AP is given then the next consecutive terms are obtained by adding the common difference on the previous term. As there are five parts of the question, we will solve each part separately.

Complete step-by-step answer:

Before solving each part of the question, first we must know what is an AP and common difference. AP is the short form of arithmetic progression or arithmetic sequence. It is a sequence of the numbers such that the difference between the consecutive terms is constant. Constant difference is the difference between the consecutive terms of AP. They are called so because these differences are constant. Now, we will solve each part of the question:
(i). \[{\rm{a}} = {\rm{1}}0\], \(d = 10\): Here the first term is 10. The second term will be obtained by adding common difference \[\left( { = {\rm{1}}0} \right)\] in the first term.\[{\rm{Second term}} = {\rm{First term}} + {\rm{1}}0 = {\rm{1}}0 + {\rm{1}}0 = {\rm{2}}0\]. Similarly the third term is obtained by adding constant difference in second term.\[{\rm{Third term}} = {\rm{ Second term}} + {\rm{1}}0 = {\rm{2}}0 + {\rm{1}}0 = {\rm{3}}0\]. The fourth term will be\[ = {\rm{Third term}} + {\rm{1}}0 = {\rm{3}}0 + {\rm{1}}0 = {\rm{4}}0\]. So the first four terms: 10, 20, 30, 40.

(ii). \[{\rm{a}} = - 2\], \(d = 0\) : Here the first term is -2. The second term will be obtained by adding common difference \[\left( { = 0} \right)\] in the first term.\[{\rm{Second term}} = {\rm{First term}} + 0 = - {\rm{2}} + 0 = - {\rm{2}}\]. Similarly\[{\rm{third term}} = {\rm{Second term}} + 0 = - {\rm{2}} + 0 = - {\rm{2}}\].\[{\rm{Fourth term}} = {\rm{Third term + }}0 = - {\rm{2}} + 0 = - {\rm{2}}\]. So the first four terms of this AP are: -2, -2, -2, -2.
(iii). \[{\rm{a}} = 4\], \(d = - 3\) : Here the first term is 4. The second term will be obtained by adding common difference \[\left( { = - 3} \right)\]in the first term.\[{\rm{Second term}} = {\rm{First term}} + \left( { - {\rm{3}}} \right) = {\rm{4}} + \left( { - {\rm{3}}} \right) = {\rm{4}} - {\rm{3}} = {\rm{1}}\]. Similarly\[{\rm{Third term}} = {\rm{Second term}} + \left( { - {\rm{3}}} \right) = {\rm{1 + }}\left( { - {\rm{3}}} \right) = {\rm{1}} - {\rm{3}} = - {\rm{2}}\].\[{\rm{Fourth term}} = {\rm{Third term}} + \left( { - {\rm{3}}} \right) = - {\rm{2}} + \left( { - {\rm{3}}} \right) = - {\rm{2}} - {\rm{3}} = - {\rm{5}}\]. So the first four terms of this AP are: 4, 1, -2, -5.

(iv). \[{\rm{a}} = - 1\] , \(d = \dfrac{1}{2}\) : Here the first term is -1. The second term will be\[ - {\rm{1 }} + {\rm{ }}\dfrac{1}{2}{\rm{ }} = {\rm{ - }}\dfrac{1}{2}\]. Similarly the third term\[ = - \dfrac{1}{2} + \dfrac{1}{2} = 0\]. Fourth term \( = 0 + \dfrac{1}{2} = \dfrac{1}{2}\) . The first four terms of this AP are: -1, \( - \dfrac{1}{2}\), 0, \(\dfrac{1}{2}\) .

(v). \[{\rm{a}} = - 1.25\], \(d = - 0.25\) : Here the first term is -1.25. The second term \( = - 1.25 + \left( { - 0.25} \right) = - 1.25 - 0.25 = - 1.50 = - 1.5\). Similarly the third\[{\rm{ = }}\left( { - {\rm{1}}.{\rm{5}}} \right) + \left( { - 0.{\rm{25}}} \right) = - {\rm{1}}.{\rm{75}}\]. Fourth term\[ = \left( { - {\rm{1}}.{\rm{75}}} \right) + \left( { - 0.{\rm{25}}} \right) = - {\rm{2}}.00 = - {\rm{2}}\]. The first four terms of this AP are: -1.25, -1.5, -1.75, -2.


Note: The terms of AP can also be calculated by the formula as shown:
\({a_n}{\rm{ = a + }}\left( {n - 1} \right)d\)
where \({a_n}\)is the nth term, a is the first term and d is the common difference. When we put the value of n = 1, 2, 3 and 4 in the above equation, we will get the first four terms of an AP.