2018/008) Five distinct 2 digit numbers are in GP. Find them

As we have $2^4=16$ and $10 * 16 > 99$ hence the common ratio $< 2$

So let the common ration be $\frac{b}{a}$ where a and b are co-primes

So the numbers are $a^4, a^3b,a^2b^2,a^3b,a^4$ and we should have

$a^4 >=10=>a>= 2$ and $b^4 < 100=> b <=3$ ands hence $a=2,b=3$ and the numbers are$16,24,36,54,81$

2018/007) consider the equation $x^4-18x^3+kx^2+174x - 2015=0$

If the product of 2 roots of eqution is -31 then find the value of k

Solution
Product of 4 roots = -2015
product of 2 roots = -31
so product of other 2 roots = -2015/(-31) =  65
so 2 quadratic factors are $x^2+ax-31$ and $x^2+bx +65$ where a and b are to be determined
so  $x^4-18x^3+kx^2+174x - 2015=(x^2+ax-31)(x^2+bx +65)$
or $x^4 + (a+b)x^3 +(65-31+ab)x^2 + (65a - 31b)x - 2015=0$
comparing coefficients we get
$a+b= -18$, $(65a-31b= 174$,$k= 34 + ab$
we can solve 1st 2 to get $a= -4, b= -14$ so putting in 3rd we get $k= 34 + ab= 90$   

2018/006) Show that if $a+b$, $c+a$, $b+c$ are in HP $a^2,b^2,c^2$ are in AP

Because $a+b$, $c+a$, $b+c$ are in HP
hence
 $\frac{1}{a+b}$, $\frac{1}{c+a}$, $\frac{1}{b+c}$ are in AP
or
 $\frac{1}{b+c}-\frac{1}{c+a}= \frac{1}{c+a}-\frac{1}{a+b}$
or
$\frac{(c+a)-(b+c)}{(b+c)(c+a)}= \frac{(a+b)-(c+a)}{(c+a)(a+b)}$
or
$\frac{a-b}{(b+c)(c+a)}= \frac{b-c}{(c+a)(a+b)}$
or $(a-b)(a+b)=(b-c)(b+c)$
or $a^2-b^2 = b^2 -c^2$
or $a^2,b^2,c^2$ are in AP