Solve the following
Question:

(i) Which term of the sequence $24,231 / 4,221 / 2,213 / 4 \ldots$ is the first negative term?

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

Solution:

(i) $24,231 / 4,221 / 2,213 / 4 \ldots$

This is an A.P.

Here, we have:

a = 24

$d=\left(23 \frac{1}{4}-24\right)=\left(-\frac{3}{4}\right)$

Let the first negative term be $a_{n}$.

Then, we have:

$a_{n}<0$

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

$\Rightarrow 24+(n-1) \quad(-3 / 4)<0$

$\Rightarrow 24-\frac{3 n}{4}+\frac{3}{4}<0$

$\Rightarrow 24+\frac{3}{4}<\frac{3 n}{4}$

$\Rightarrow \frac{99}{4}<\frac{3 n}{4}$

$\Rightarrow 99<3 n$

$\Rightarrow n>33$

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.

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

$\Rightarrow n=5$

(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.

$\therefore(13-n)=0$

$\Rightarrow n=13$

Solve the following
Question:

If $\left(1-x+x^{2}\right)^{n}=a_{0}+a_{1} x+a_{2} x^{2}+\ldots+a_{2 n} x^{2 n}$, find the value of $a_{0}+a_{2}+a_{4}+\ldots+a_{2 n}$.

Solution:

Putting x = 1 and −1 in

$\left(1-x+x^{2}\right)^{n}=a_{0}+a_{1} x+a_{2} x^{2}+\ldots+a_{2 n} x^{2 n}$

We get,

$1=a_{0}+a_{1}+a_{2}+\ldots+a_{2 n} \quad \ldots(1)$

And,

$3^{n}=a_{0}-a_{1}+a_{2}-\ldots+a_{2 n} \quad \ldots(2)$

Adding (1) and (2), we get

$3^{n}+1=2\left(a_{0}+a_{2}+\ldots+a_{2 n}\right)$

Hence, the value of $a_{0}+a_{2}+a_{4}+\ldots+a_{2 n}$ is $\frac{3^{n}+1}{2}$.