Ques1) Find the sum of the series
1/9 + 1/18 + 1/30 + 1/45 + 1/63 + .....
(a) 1/3 (b) 1/4 (c) 1/5 (d) 2/3
Ques2) There exisits positive integers A, B and C with no common factors greater than 1, such that A log200 5 +B log200 2 = C. The sum A+B+C equlas
(a) 5 (b) 6 (c)7 (d) 8
Ques3) The sequence a1,a2,a3,..... satisfies a1=19, a9=99 and for all n≥3 , an is the arithmetic mean of the first n-1 terms. Then a2 is equals to
(a) 179 (b)99 (c)79 (d) 59
Ques4) If b is the arithmetic mean b/w a and x; b is the geometric mean b/w a and y; b is the harmonic mean b/w a and z, (a,b,x,y,z >0), then the value of xyz is
(a) a3 (b) b3 (c) {b3(2a-b)} / (2b-a) (d) { b 3 (2b-a)}/ (2a-b)
Ques5) If a,b,c are numbers for which the equation
{x2 +10x -36} / x(x-3)2 = a/x + b/(x-3) + c /(x-3)2 is an identity , then a+b+c equals to
(a) 2 (b) 3 (c) 10 (d)8
106Q5. \frac{x^2+10x-36}{x(x-3)^2} = \frac{a}{x}+\frac{b}{x-3}+\frac{c}{(x-3)^2}
taking LCM and multiplying
\frac{x^2+10x-36}{x(x-3)^2} = \frac{a(x-3)^2+bx(x-3)+cx}{x(x-3)^2}
simplify and equate the coefficients of x2,x and the common term
1=a+b, 10= -6a-3b+c and -36 = 9a will be the three equations
from last equation, a= -4, so b=5 and hence c=1
This method is called partial fractions which is used in integration as well
106Q4. If b is the arithmetic mean b/w a and x; b is the geometric mean b/w a and y; b is the harmonic mean b/w a and z, (a,b,x,y,z >0), then the value of xyz is
(a) a3 (b) b3 (c) {b3(2a-b)} / (2b-a) (d) { b 3 (2b-a)}/ (2a-b)
2b=x+a ... (i) ==> x = (2b-a)
b2=ay ... (ii) ==> y=b2/a
2/b=1/a+1/z ... (iii) ==> z = ab/(2a-b)
i hope u can proceed from here now to get answer as (d)
112) is quite easy, get it in the form 5A2B=52C23C, since (A,B) are co-prime, C=1, thus A+B+C=6...
621/9 + 1/18 + 1/30 + 1/45 + 1/63 + .....
1/3\left\{1/3+1/6+1/10+1/15+1/21....\right\}
look at 3, 6, 10, 15, 21
Their first difference is in AP..
now can you solve :)
11Observe that all the terms are of the form 19+a22=99, thus a2 can be got = 179...
