Mathematics Test-38

Given,

The sum of the 3rd and the 7th term of an A.P. is 30 and the sum of the 5th and the 9th term is 56.

Let the first term of the A.P. be 'a' and the common difference 'd'.

nth term of an A.P.,

\(a_n = a + (n-1)d\)

\(a_3 = a + (3-1)d\)

\(\Rightarrow a_3 = a + 2d\)

\(a_7 = a + (7-1)d\)

\(\Rightarrow a_7 = a + 6d\)

\(a_5 = a + (5-1)d\)

\(\Rightarrow a_7 = a + 4d\)

\(a_9 = a + (9-1)d\)

\(\Rightarrow a_9 = a + 8d\)

Therefore, we can write

\({a}_{3}+a_{7}\) = 30

a + 2d + a + 6d = 30

2a + 8d = 30 ...(i)

And, \({a}_{5}+a_{9}\) = 56

a + 4d + a + 8d = 56

2a + 12d = 56 ...(ii)

On solving equation (i) and equation (ii), we get

4a + 20d = 86

2a + 10d = 43.....(iii)

\({a}_{4}+a_{8}\) = a + (4-1)d + a + (8-1)d

= a + 3d +a + 7d

= 2a +10d

From equation (i), we get

The sum of the fourth and eighth terms of the series,

\({a}_{4}+a_{8}\) = 43

Given,

\(A=\left[\begin{array}{ccc}1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1\end{array}\right], B=\left[\begin{array}{ccc}3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3\end{array}\right]\)

\(\therefore A+B=\left[\begin{array}{ccc}1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1\end{array}\right]+\left[\begin{array}{ccc}3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3\end{array}\right]\)

\(\Rightarrow \mathrm{A}+\mathrm{B}=\left[\begin{array}{ccc}1+3 & 2-1 & -3+2 \\ 5+4 & 0+2 & 2+5 \\ 1+2 & -1+0 & 1+3\end{array}\right]\)

\(\Rightarrow \mathrm{A}+\mathrm{B}=\left[\begin{array}{ccc}4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4\end{array}\right]\)

As we know that the point \((x_1, y_1, z_1)\) of \(a x+b y+c z+d=0\) is,

\(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=\frac{-2\left(a x_1+b y_1+c z_1+d\right)}{a^2+b^2+c^2}\)

So the point \((-1,3,4)\) in the plane \({x}-2 {y}=0\) is given by,

\(\frac{x+1}{1}=\frac{y-3}{-2}=\frac{z-4}{0}=\frac{-2[1(-1)+(-2(+3) )+0(4)]}{1+4}\)

\(\frac{x+1}{1}=\frac{y-3}{-2}=\frac{z-4}{0}=\frac{-2(-7)}{5}\)

\(x=\frac{14}{5}-1=\frac{9}{5}\)

\(y=\frac{-28}{5}+3=\frac{-13}{5}\) and \(z=4\)

The point is \((\frac{9}{5}, \frac{-13}{5}, 4)\).

Given determinant is \(\left|\begin{array}{lll}0 & c & b \\ c & 0 & a \\ b & a & 0\end{array}\right|^{2}\), simplifying the determinant based on first row we get,

\(=\left\{0 \times\left(0-a^{2}\right)-c \times(c \times 0-b \times a)+b(c \times a-b \times 0)\right\}^{2}\)

\(=(a b c+a b c)^{2}\)

\(=4 a^{2} b^{2} c^{2}\)

So, the right answer is \(4 \mathrm{a}^{2} \mathrm{~b}^{2} \mathrm{c}^{2}\)

Let \(z=x+i y=(-1-i \sqrt{3})\)

So, \(x=-1\) and \(y=-\sqrt{3}\)

Here sign of \(x\) and \(y\) are negative so \(z\) lies in third quadrant.

Now,

Principal argument of \((-1-\mathrm{i} \sqrt{3})=-\pi+\tan ^{-1}\left|\frac{\mathrm{y}}{\mathrm{x}}\right|\)

\(=-\pi+\tan ^{-1}\left|\frac{\sqrt{3}}{1}\right| \)

\(=-\pi+\frac{\pi}{3} \)

\(=-\frac{2 \pi}{3}\)

As we know,

General term in the expansion of \((x+y)^{n}\) is given by,

\(T _{( r +1)}={ }^{ n } C _{ r } \times x ^{ n - r } \times y ^{ r }\)

The middle term is the expansion of \((x+y)^{n}\) depends upon the value of \(n\).

If \(n\) is even, then total number of terms in the expansion of \((x+y)^{n}\) is \(n+1\).

So there is only one middle term i.e., \(\left(\frac{n}{2}+1\right)\)th term.

If \(n\) is odd, then total number of terms in the expansion of \(( x + y )^{ n }\) is \(n +1\).

So there are two middle terms i.e., \(\left(\frac{ n +1}{2}\right)^{\text {th }}\) and \(\left(\frac{ n +3}{2}\right)^{\text {th }}\).

Here \(n=5\) ( \(n\) is odd number)

\(\therefore\) Middle term \(=\left(\frac{ n +1}{2}\right)^{\text {th }}\) and \(\left(\frac{ n +3}{2}\right)^{\text {th }}=3^{\text {rd }}\) and \(4^{\text {th }}\)

\(\therefore T _{3}= T _{(2+1)}={ }^{5} C _{2} \times(2 x )^{(5-2)} \times\left(\frac{1}{ x }\right)^{2}\) and \(T _{4}= T _{(3+1)}={ }^{5} C _{3} \times(2 x )^{(5-3)} \times\left(\frac{1}{ x }\right)^{3}\)

\(\Rightarrow T _{3}={ }^{5} C _{2} \times\left(2^{3} x \right)\) and \(T _{4}={ }^{5} C _{3} \times 2^{2} \times \frac{1}{ x }\)

\(\Rightarrow T _{3}=80 x\) and \(T _{4}=\frac{40}{ x }\)

So, the middle terms in the expansion of \(\left(2 x +\frac{1}{ x }\right)^{5}\) are \(80 x\) and \(\frac{40}{x}\).

\(\angle \mathrm{APB}=50^{\circ}\) (Given)

By degree measure theorem: \(\angle A O B=2 \angle A P B\)

\(\angle \mathrm{AOB}=2 \times 50^{\circ}=100^{\circ}\)

Again, \(\mathrm{OA}=\mathrm{OB}\) [Radius of circle]

Then \(\angle O A B=\angle O B A\) [Angles opposite to equal sides]

Let \(\angle O A B=m\)

In \(\triangle O {A B}\)

By angle sum property: \(\angle O A B+\angle O B A+\angle A O B=180^{\circ}\)

\(\Rightarrow m+m+100^{0}=180^{\circ}\)

\(\Rightarrow2 \mathrm{~m}=180^{\circ}-100^{\circ}=80^{\circ}\)

\(\Rightarrow \mathrm{m}=\frac{80^{\circ}}{2}=40^{\circ}\)

\(\angle O A B=\angle O B A=40^{\circ}\)