Question

(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 $12+8i,11+6i,10+4i,\dots \dots$ is (a) purely real (b) purely imaginary ?

Answer 1

The given sequence is $24,23\frac{1}{4},22\frac{1}{2},22\frac{3}{4},\dots \dots$

Here, a = 24

$d=23\frac{1}{4}-24=\frac{93-96}{4}=\frac{-3}{4}$

Let nth term be the 1st negative term.

$a_n<0$

$\Rightarrow a+(n-1)d<0$

$\Rightarrow 24-\frac{3}{4}(n-1)<0$

$\Rightarrow 96-3n+3<0$

$\Rightarrow 99<3n$

$\Rightarrow 33<nOrn>33$

$\therefore$ 34th term is 1st negative term.

(ii) (a) 12 + 8i, 11 + 6i, 10 + 4i, ......

This is an A.P.

Here, we have:

a = 12 + 8i

d = (11 + 6i - 12 - 8i)

= (-1 - 2i)

Let the real term be $a_n=a+(n-1)d$

$a_n=(12+8i)+(n-1)(-1-2i)$

$=(12+8i)+(-n+1-2in+2i)$

$=12+8i-n+1-2in+2i$

$=(13-n)+(8-2n+2)i$

$=(13-n)+(10-2n)i$

$a_n$ has to be real.

$\therefore (10-2n)=0$

$\Rightarrow n=5$

(b) 12 + 8i, 11 + 6i, 10 + 4i, .....

This is an A.P.

Here, we have:

a = 12 + 8i

d = (11 + 6i - 12 - 8i)

= (-1 -2i)

Let the imaginary term be $a_n=a+(n-1)d$

$a_n=(12+8i)+(n-1)(-1-2i)$

$=(12+8i)+(-n+1-2in+2i)$

$=12-8i-n+1-2in+2i$

$=(13-n)+(8-2n+2)i$

$=(13-n)+(10-2n)i$

$a_n$ has to be imaginary.

$\therefore (13-n)=0$

$\Rightarrow n=13$