(i) Which term of the sequence 24, $23 \frac{1}{4,} 22 \frac{1}{2,} 21 \frac{3}{4}$ ... is the first negative term?
(ii) Which term of the sequence 12 + 8i, 11 + 6i, 10 + 4i, ... is (a) purely real (b) purely imaginary?

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Solution

(i) 24, $23 \frac{1}{4,} 22 \frac{1}{2,} 21 \frac{3}{4}$ ...
This is an A.P.
Here, we have:
a = 24
$d = (23 \frac{1}{4} - 24) = (- \frac{3}{4}) Let the first negative term be a_{n} . Then, w e h a v e : a_{n} < 0 \Rightarrow a + (n - 1) d < 0 \Rightarrow 24 + (n - 1) (- \frac{3}{4}) < 0 \Rightarrow 24 - \frac{3 n}{4} + \frac{3}{4} < 0 \Rightarrow 24 + \frac{3}{4} < \frac{3 n}{4} \Rightarrow \frac{99}{4} < \frac{3 n}{4} \Rightarrow 99 < 3 n \Rightarrow n > 33$

Thus, the 34th term is the first negative term of the given A.P.

(ii)
(a) 12 + 8i, 11 + 6i, 10 + 4i...
This is an A.P.
Here, we have:
a = 12 + 8i
$d = (11 + 6 i - 12 - 8 i) = (- 1 - 2 i) Let the real term be a_{n} = a + (n - 1) d . a_{n} = (12 + 8 i) + (n - 1) (- 1 - 2 i) = (12 + 8 i) + (- n + 1 - 2 i n + 2 i) = 12 + 8 i - n + 1 - 2 i n + 2 i = (13 - n) + (8 - 2 n + 2) i = (13 - n) + (10 - 2 n) i a_{n} has to be real . ∴ (10 - 2 n) = 0 \Rightarrow n = 5$

(b) 12 + 8i, 11 + 6i, 10 + 4i...
This is an A.P.
Here, we have:
a = 12 + 8i
$d = (11 + 6 i - 12 - 8 i) = (- 1 - 2 i) Let the imaginary term be a_{n} = a + (n - 1) d a_{n} = (12 + 8 i) + (n - 1) (- 1 - 2 i) = (12 + 8 i) + (- n + 1 - 2 i n + 2 i) = 12 + 8 i - n + 1 - 2 i n + 2 i = (13 - n) + (8 - 2 n + 2) i = (13 - n) + (10 - 2 n) i a_{n} has to be imaginary . ∴ (13 - n) = 0 \Rightarrow n = 13$

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Similar questions

(i) Which term of the sequence $24, 23 \frac{1}{4}, 22 \frac{1}{2}, 22 \frac{3}{4}, \dots \dots$ is the first negative term ?

(ii) Which term of the sequence $12 + 8 i, 11 + 6 i, 10 + 4 i, \dots \dots$ is (a) purely real (b) purely imaginary ?

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