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Answer :
(i) a = 7, d = 3, n = 8
We need to find \(a_n\) here.
Using formula \(a_n = a + (n - 1)d\)
Putting values of a, d and n,
\( a_n\) = 7 + (8 - 1) 3
= 7 + (7) 3 = 7 + 21 = 28
(ii) a = –18, n= 10, \(a_n = 0\)
We need to find d here.
Using formula \(a_n = a + (n - 1)d\)
Putting values of a,\(a_n , n\)
\(\Rightarrow \) 0 = –18 + (10 – 1) d
\(\Rightarrow \) 0 = -18 + 9d
\(\Rightarrow \) 18 = 9d
\(\Rightarrow \) d = 2
(iii) d = –3, n = 18, \(a_n = -5\)
We need to find a here.
Using formula \(a_n = a + (n - 1)d\)
Putting values of d, \(a_n , n\)
\(\Rightarrow \) –5 = a + (18 – 1) (–3)
\(\Rightarrow \) - 5 = a + (17) (-3)
\(\Rightarrow \) -5 = a – 51
\(\Rightarrow \) a = 46
(iv) a = –18.9, d = 2.5, \(a_n = 3.6\)
We need to find n here.
Using formula \(a_n = a + (n - 1)d\)
Putting values of \(d, a_n , n\)
\(\Rightarrow \) 3.6 = –18.9 + (n – 1) (2.5)
\(\Rightarrow \) 3.6 + 18.9 = (n - 1)(2.5)
\(\Rightarrow \) 22.5 = (n -1)(2.5)
\(\Rightarrow \) (n - 1) = \(\frac{22.5}{2.5}\)
\(\Rightarrow \) n - 1 = 9
\(\Rightarrow \) n = 10
(v) a = 3.5, d = 0, n = 105
We need to find \(a_n\) here.
Using formula \(a_n = a + (n - 1)d\)
Putting values of d, n and a,
\(\Rightarrow a_n\) = 3.5 + (105 - 1) (0)
\(\Rightarrow \)\(a_n = 3.5 + 104 × 0\)
\(\Rightarrow \) \(a_n = 3.5 + 0 \)
\(\Rightarrow a_n = 3.5\)