Write first four terms of the AP, when the first term a and the common difference d are given as follows: (i) a = 10, d = 10 (ii) a = –2, d = 0 (iii) a = 4, d = – 3 (iv) a = – 1, d = 1 2 (v) a = – 1.25, d = – 0.25
Step-by-Step Solution
Key Concept: In an arithmetic progression (AP) the nth term is given by \(a_n = a + (n-1)d\), where \(a\) is the first term and \(d\) is the common difference. The first four terms correspond to \(n = 1,2,3,4\).
1. Write the general formula for the nth term of an AP: \[a_n = a + (n-1)d.\]
2. Substitute \(n = 1,2,3,4\) to obtain the first four terms:
- \(a_1 = a + 0\cdot d = a\)
- \(a_2 = a + 1\cdot d = a + d\)
- \(a_3 = a + 2\cdot d = a + 2d\)
- \(a_4 = a + 3\cdot d = a + 3d\)
3. For each given pair \((a,d)\), compute the four terms using the above expressions.
(i) \(a = 10, d = 10\):
\(a_1 = 10\)
\(a_2 = 10 + 10 = 20\)
\(a_3 = 10 + 2\times10 = 30\)
\(a_4 = 10 + 3\times10 = 40\)
(ii) \(a = -2, d = 0\):
\(a_1 = -2\)
\(a_2 = -2 + 0 = -2\)
\(a_3 = -2 + 0 = -2\)
\(a_4 = -2 + 0 = -2\)
(iii) \(a = 4, d = -3\):
\(a_1 = 4\)
\(a_2 = 4 + (-3) = 1\)
\(a_3 = 4 + 2(-3) = -2\)
\(a_4 = 4 + 3(-3) = -5\)
(iv) \(a = -1, d = \frac{1}{2}\):
\(a_1 = -1\)
\(a_2 = -1 + \frac{1}{2} = -\frac{1}{2}\)
\(a_3 = -1 + 2\times\frac{1}{2} = 0\)
\(a_4 = -1 + 3\times\frac{1}{2} = \frac{1}{2}\)
(v) \(a = -1.25, d = -0.25\):
\(a_1 = -1.25\)
\(a_2 = -1.25 + (-0.25) = -1.50\)
\(a_3 = -1.25 + 2(-0.25) = -1.75\)
\(a_4 = -1.25 + 3(-0.25) = -2.00\)
4. List the results for each part.
Correct Answer: {"(i)":[10,20,30,40],"(ii)":[-2,-2,-2,-2],"(iii)":[4,1,-2,-5],"(iv)":[-1,-0.5,0,0.5],"(v)":[-1.25,-1.5,-1.75,-2.0]}