2016/092) a,b,c, are 1st 3 terms of a geometric series . if harmonic mean of a,b is 12 and of b,c is 36 find first 5 terms of the series.

let common ratio be t
then terms are a, at
the harmonic mean is 12
so $\frac{1}{a} + \frac{1}{at} = \frac{1}{6}\cdots(1)$
and
$\frac{1}{at} + \frac{1}{at} = \frac{1}{18}$
deiding (1) by (2) we get $t = 3$
now from (1) $\frac{1}{a} + \frac{1}{3a} = \frac{1}{6}$ or a = 8
giving 5 terms 8,24,72,216,648.

2016/91) if $a_1,a_2,\cdots,a_n$ are in HP show that $a_1a_2+a_2a_3+\cdots + a_{n-1}a_n = (n-1)a_1a_n$

we have
  $\frac{1}{a_1},\frac{1}{a_2}\cdots,\frac{1}{a_n}$ are in AP hence
or $(n-1)(\frac{1}{a_2} - \frac{1}{a_1})  = \frac{1}{a_n} - \frac{1}{a_1}$
or $(n-1)(a_1-a_2)(a_n a_1) = a_2a_1(a_1-a_n) \cdots(1)$
similarly
 $(n-1)(a_2-a_3)(a_n a_1) = a_3a_2(a_1-a_n) \cdots(2)$\
...
   $(n-1)(a_{n-1}-a_n)(a_n  a_1) = a_{n-1}a_m(a_1-a_n) \cdots(n-1)$\
adding above n eqautions we get
$(n-1)(a_1-a_n)(a_n a_1) = (a_1a_2+a_2a_3+\cdots + a_{n-1}a_n) (a_1-a_n)$
cancelling $a_1-a_n$ from both sides we get the result

2016/090) Find the value of $\frac{18^3+7^3+3.18 *7 *25}{3^6+6*243 *2 + 15 *81 *4+20 *27 *8 + 15 *9 *16 + 6*3*32 + 64}$

We have numerator = $18^3+7^3+3.18*7*25 = 18^3+7^3 + 3 *18 *7 (18+7) = (18+7)^3 = 25^3 = 5^6$
denominator = $3^6+6*243*2 + 15*81*4+20*27*8 + 15*9*16 + 6*3*32 + 64$*
$= 3^6 + 6 * 3^5 * 2 + 15 * 3^4 * 2^2  + 20 * 3^3 *2^3 + 15 * 3^2 * 2^4 + 6 * 3 * 2^5 + 2^6$
$= {6 \choose 0}3^6 + {6 \choose 1} * 3^5 * 2 + {6 \choose 2} * 3^4 * 2^2  + {6 \choose 3} * 3^3 *2^3 + {6 \choose 4} * 3^2 * 2^4 + {6 \choose 5} * 3 * 2^5 + {6 \choose 6}2^6$
$= ( 3+2)^6 = 5^6$
so given expression is 1

2016/089) Evaluate $\sin\,18^\circ$ and $\cos\,18^\circ$

Let, $A = 18^\circ$
Then $2A = 90^\circ - 3A$
Taking sine on both sides, we get
$\sin 2A = \sin (90^\circ - 3A) = \cos 3A$
$=> 2 \sin\, A \cos\, A = 4 \cos^3 A - 3 \cos\, A$
or $2 \sin\, A \cos\, A - 4 \cos^3 A + 3 \cos\, A = 0$
or $\cos\, A (2 \sin\, A - 4 \cos^2 A + 3) = 0$
Dividing both sides by $\cos\, A = \cos 18^\circ$ which is not zero we get
$2 \sin\, A - 4 (1 - \sin^2 A) + 3 = 0$
or $4 \sin^2 A + 2 \sin\ A - 1 = 0$ which is a quadratic in $\sin\ A$
hence $\sin\,A = \frac{-1\pm\sqrt{5}}{4}$
but as $\sin\, 18^\circ$ is positive we have $\sin 18^\circ = \frac{-1+\sqrt{5}}{4}$
now $\cos^2 18^\circ= 1- \sin ^2 18^\circ = 1 - (\frac{-1+\sqrt{5}}{4})^2$
$= 1 - \frac{5+1-2\sqrt{5}}{16} = \frac{10+2\sqrt{5}}{16}$
 $\cos\,18^\circ= \frac{\sqrt{10+2\sqrt{5}}}{4}$

2016/088) Which number is smaller $\sqrt{3} + \sqrt{5}$ or $\sqrt{2} + \sqrt{6}$

we have $\sqrt{3} - \sqrt{2}= \frac{1}{\sqrt{3} + \sqrt{2}}$
and  $\sqrt{6} - \sqrt{5}= \frac{1}{\sqrt{6} + \sqrt{5}}$
from the above $\sqrt{3} - \sqrt{2} > \sqrt{6} - \sqrt{5}$
or  $\sqrt{3} + \sqrt{5} > \sqrt{6} + \sqrt{2}$

2016/087) If $a\sin\,x=b\sin(x+\frac{2\pi}{3})=c\sin(x+\frac{4\pi}{3})$ prove that $ab+bc+ca=0$

Let $asin\,x=b\sin(x+\frac{2\pi}{3})=c\sin(x+\frac{4\pi}{3})=k$
hence $\frac{k}{a} = \sin\,x\cdots(1)$
$\frac{k}{b} = \sin(x+\frac{2\pi}{3})\cdots(2)$
$\frac{k}{c} = \sin(x+\frac{4\pi}{3})$
or $\frac{k}{c} = \sin(x-\frac{2\pi}{3})\cdots(3)$

from (1),(2) and (3)
$\frac{k}{a} + \frac{k}{b} + \frac{k}{c} =\sin\,x +  \sin(x+\frac{2\pi}{3}) + \sin(x-\frac{2\pi}{3})$
$= \sin\,x +  \sin\,x\cos \frac{2\pi}{3} + \cos \,x\sin  \frac{2\pi}{3} + \sin\,x\cos \frac{2\pi}{3} - \cos \,x\sin  \frac{2\pi}{3}$
$= \sin\,x +  2\sin\,x\cos \frac{2\pi}{3}$
$= \sin\,x +  2\sin\,x( -\frac{1}{2})$
$= \sin\,x -  \sin\,x$
$= 0$

2016/086) If a,b,c are in AP find the fixed point wthough which line $ax+by+c= 0$ passes

a,b,c are in AP so $a+c = 2b$ or $c = 2b-a$
$ax+by+c=0$
$=>ax + by + (2b-a)=0$ or $(x-1) a + (y+2) b=0$
so the point through which the lines pass is (1,-2) as the above equation should be independent of (a,b)