2015/062) A problem in AP

There are 2 sets of numbers each consisting of 3 terms in A.P & sum of each set is 15. The common difference of the $1^{st}$ set is greater than the common difference of the $2^{nd}$ set by 1 and the ratio of product of the $1^{st}$ set is to the product of $2^{nd}$ set is 7 to 8.Find the numbers. 

Solution
 
As sum of 3 terms is 15 so middle term is 5


let the common difference in $1^{st}$ set be a, so common difference in $2^{nd}$ set is (a-1)

then numbers in 1^st set are 5-a, 5, 5+ a and in 2^nd set are the numbers are 6-a, 5, 4+ a

now as per given condition $\dfrac{(5-a)5(5+a)}{((6-a) 5 (4+a))} = \dfrac{7}{8}$

or $8(25-a^2)= 7(6-a)(4+a) = 7(24+ 2a - a^2)$

or $200- 8a^2 = 168 + 14a - 7a^2$ or $a^2 + 14 a - 32 = 0$

$(a-2)(a+16) = 0$

a= 2 gives a- 1 = 1 gives 1st series = 3,5,7 and second series = 4,5,6

a = - 16 gives a- 1= - 17 the 1st series = 21,5, -11 and second series = 22,5, - 12.

refer to http://in.answers.yahoo.com/question/index;_ylt=Ar2kcwS_d26mHg5f2W.h9F.RHQx.;_ylv=3?qid=20130222042246AAW7rSk

2015/061) Simplify $((-1)+\frac{i\sqrt{2}}{3})^2 + ((-1)-\frac{i\sqrt{2}}{3})^2$

we know

$(a+b)^2 + (a-b)^2 = 2 (a^2+b^2)$

gives $((-1)+\frac{i\sqrt{2}}{3})^2 + ((-1)-\frac{i\sqrt{2}}{3})^2$

  =$2((-1)^2+(\frac{i\sqrt{2}}{3})^2)$

 $= 2 ( 1- \frac{2}{9}) = \frac{14}{9}$

2015/060) If the roots of the equation

$(a²+b²)x²-2(ac+bd)x+(c²+d²)=0$ are equal, prove that $\dfrac{a}{b}=\dfrac{c}{d}$

Solution

For roots to be equal, discriminant = 0
$=> [-2(ac + bd)]^2 - 4 * (a^2 + b^2) (c^2 + d^2) = 0$
$=> 4[(ac)^2 + 2abcd + (cd)^2] - 4[(ac)^2 + (ad)^2 + (bc)^2 + (cd)^2] = 0$
$=> [(ac)^2 + 2abcd + (cd)^2] - [(ac)^2 + (ad)^2 + (bc)^2 + (cd)^2] = 0$
$=> (2abcd - (bc)^2 - (ad)^2) = 0$
$=> (bc)^2 - 2abcd + (ad)^2 = 0$
$=> (bc-ad)^2 = 0$
$=> bc = ad$
$=> \dfrac{a}{b} = \dfrac{c}{d}$

2015/059) Problem Find the largest number

N and M are positive integers such that N+M=21. The largest possible value of $\dfrac{1}{N}+\dfrac{1}{M}$ is $\dfrac{a}{b}$ where $\dfrac{a}{b}$ is in lowest form.Find a+b.

Solution

$\dfrac{1}{N}+\dfrac{1}{M}$ = $\dfrac{N+M}{NM}$ =$\dfrac{21}{NM}$

This is largest when NM is lowest.

clearly when N= 1 and M= 20 ( or N = 20, M = 1)

so NM = 20

so a= 21 and b = 20 hence a+b = 41

I have solved at https://in.answers.yahoo.com/question/index?qid=20130408091854AAlkBX4

 

2015/058) 100 positive integers are written in a row. The average of the first and second numbers is 1. The average of the second and third numbers is 2. The average of the third and fourth numbers is 3. This pattern continues, and the average of the 99th and 100th number is 99. What is the 100th number?

Average of $1^{st}$ and $2^{nd}$ number is 1 so sum is 2 so both positive integers have to be 1

so $a_1= 1$
$a_2 = 1$

now average of $2^{nd}$ and $3^{rd}$ is 2 so sum = 4 so $a_3= 3$

average of $3^{rd}$ and $4^{th}$ is 3 so sum = 6 so $a_4= 3$

average of $4^{th}$ and $5^{th}$ is 4 so sum = 6 so $a_5= 5$

so we have $n^{th}$ term is n when n is odd and n-1 when n is even

this can be proved as below

for n odd 2 numbers are n and n and average = n
for n even the number is n-1 and next number is n+1 and average is n

so $100^{th}$ number = 99

refer to https://in.answers.yahoo.com/question/index?qid=20121112041716AA4V52E

2015/057) Three roots of the equation $4p^3 - 3p - 0.5 = 0$ all lie between 1 and -1. Solve

If we let $p = \cos,t$ then we get

$\cos 3t = .5 = \cos \dfrac{\pi}{3}$

so we get $3t$ =$\dfrac{\pi}{3}$ or $\dfrac{7\pi}{3}$ or $\dfrac{13\pi}{3}$

or  $t$ =$\dfrac{\pi}{9}$ or $\dfrac{7\pi}{9}$ or $\dfrac{13\pi}{9}$

or $p$ =  $\cos \dfrac{\pi}{9}$ or $\cos \dfrac{7\pi}{9}$ or $\cos \dfrac{13\pi}{9}$

2015/056) How many ordered pairs of integers $(a,b)$ satisfy $\dfrac{1}{a} +\dfrac{1}{b}= \dfrac{1}{6}$

 $\dfrac{1}{a} +\dfrac{1}{b}= \dfrac{1}{6}$

hence $\dfrac{a+b}{ab}= \dfrac{1}{6}$

or $6 * (b + a) = ab$
or $6b + 6a = ab$

or $ab – 6b – 6a + 36 = 36$

or $(a-6)(b-6) = 36$
So  b - 6 must be a divisor of 36

Divisors:

-36 , -18 , -12 , -9 , -6 , -4 , -3 , -2 , -1 , 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36

except that b cannot be 0 as ( or b- 6 cannot be -6) as b is in denominator

secondly we have ordered pair so (a,b) and (b,a) are different

so number of ordered pairs = 17 ( one for each b -6 from -36 , -18 , -12 , -9 , -6 , -4 , -3 , -2 , -1 , 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36 except b-6 = -6 )
This I have solved at https://in.answers.yahoo.com/question/index?qid=20130530212220AA8Jqq3

2015/055) Prove that equation $(3x + 4y)(4x + 5y) = 7^z$ is not possible?

we should have

$7^a$ divides $(3x+4y)$ and $7^b$ divides $(4x+5y)$

now $gcd(3x+4y, 4x+ 5y)$
$= gcd( 3x + 4y, x+ y)$
$= gcd( y, x+y)$
$= gcd(x,y)$

if 7 divides $3x + 4y$ then it does not divide $4x+ 5y$ or if it divides both x an y then sum cannot be power of 7 hence it is not possible
This I have solved at https://in.answers.yahoo.com/question/index?qid=20130925091522AAvX8Z0

2015/054) What is the equation of a line perpendicular to $5x-6y=30$

The slope of the line $5x-6y = 30$ say $m_1$  is $\frac{5}{6}$
so slope of the line perpendicular to it say $m_2$ is  $\frac{-6}{5}$ as $m_1m_2 = -1$

so equation of line
= $6x + 5y = c# where c is a constant

as y intercept = 2 so y = 2 when x = 0
so it is $6x + 5y = 10$ is the equation

2015/053) given |a | < 1 and |b| < 1 If $1+a+a^2+\cdots=x$ and $1+b+b^2+\cdots =y$ then find $1+ab+a^2b^2+\cdots$ in terms of x and y

we have
$x = \dfrac{1}{1-a}$
or $(1-a) = \dfrac{1}{x}$ or $a = 1 – \dfrac{1}{x} = \dfrac{x-1}{x}$
similarly
$b = \dfrac{y-1}{y}$
now as |a | and |b| both < 1 so |ab| < 1 and hence
$1+ ab + a^2b^2 + a^3b^3 + \cdots= \dfrac{1}{1-ab} = \dfrac{1}{1- \frac{x-1}{x} * \frac{y-1}{y}}$
= $\dfrac{xy}{xy - (x - 1)(y – 1)}$
= $\dfrac{xy}{x + y – 1}$