Mathametics 01 – Algebra

1)  If a, b , c, d and p are distinct real numbers such that (a2 + b2 + c2) p2 – 2(ab + bc + cd ) p + (b2 + c2 + d2) ≤ 0 , then a, b, c, d are in

(a) A.P.           (b) G.P.            (c) H. P.            (d) none of these

(b) The given inequality can be re-written as (ap – b)2 + (bp – c)2 +( cp – d)2 ≤ 0 

⇒ap – b = bp – c = cp – d = 0 

⇒ a/b = b/c = c/d = 1/p 

a, b, c, d are in G. P.

2) In an infinite geometric series, with first term a and common ratio r, the second term is 3/4 and its sum is 4. Then

(a) a = 7/4 , r = 3/7               (b) a = 3/2 , r = 1/2                (c) a = 2, r = 3/8              (d) a = 3, r = 1/4

(d) Here, the sum of the series = 4 = a/(1-r) and the second term = 3/4 = ar

⇒4 = 3/[4r(1-r)]

16r2 – 16r + 3 = 0

⇒(4r – 1) (4r – 3) = 0

⇒ r = 1/4 ⇒ a = 3.

3) The sum of the first n terms of the series 12+2.22+32+2.42+52+2.62+ . . . is n(n+1)2/2, when n is even. When n is odd, the sum is

(a) [n2(n+1)]/2            (b) [n(n+1)(2n+1)]/6                (c) [n(n+1)2]/2             (d) [n2(n+1)2]/2

(a) If n is odd, n-1 is even. Sum of (n-1) terms will be [(n-1)(n-1+1)2]/2 = [n2(n-1)]/2.

The nth term will be n2. Hence the required sum = [n2(n-1)]/2+n2 = [n2(n+1)]/2 .

4) Let α, β be the roots of x2 – x + p = 0 and γ, δ be the roots of x2 – 4x + q = 0. If α, β, γ, δ are in G.P., then the integral values of p and q respectively are

(a) -2,-32          (b) -2,3               (c) -6,3             (d) -6,-32

(a) Here α + β = 1, αβ = p, γ + δ = 4, δγ = q. Since, α, β, γ, δ are in G.P., with common ratio (say) r,

α (1 + r) = 1, α (r2 + r3) = 4

⇒ r2 = 4 ⇒ r = – 2.

⇒ α = – 1 for r = -2 (for integral values of p, q) and

p = αβ = α2r = -2,

q = γδ = α2r5 = – 32

5) If p, q, r are in A.P. , then pth, qth and rth terms of any G.P. are in

(a) A.P          (b) G.P               (c) H.P             (d) A.G.P

(b) Let the first term of a G.P. be A and common ratio be R .Then pth, qth and rth terms are ARp-1, ARq-1 and ARr-1.

Obviously

ARp-1×ARr-1 = A2Rp+r-2 = (ARq-1)2,

as p + r = 2q. Hence terms are in G.P.

6) If 4a2 + 9b2 + 16c2 = 2(3ab + 6bc + 4ca), where a, b, c are non-zero numbers, then a, b, c are in

(a) A.P.                (b) G.P.                (c) H.P                 (d) none of these

(c) The given equation 4a2 + 9b2 + 16c2 = 2(3ab + 6bc + 4ca)

⇒ (2a – 3b)2 + (3b – 4c)2 + (4c – 2a)2 = 0

⇒ 2a = 3b = 4c

⇒ a/(1/2) =  b/(1/3) = c/(1/4) ⇒ a, b, c are in H.P.

        n

7) If  log(ar/br-1)=klog(an+1/bn-1), then k is equal to

       r=1   

(a) n+1                (b) n                (c) n/2                 (d) (n+1)/2

(c) We have

 n                         n 

log(ar/br-1)= log(ar-logbr-1)

r=1                   r=1

   n

= (rlog-(r-1)log b)

  r=1

=[n(n+1)]/2 log a-[(n-1)n]/2 log b

 

=n/2[(n+1)log a-(n-1)log a]

 

=n/2[log an+1/bn-1]

                                         ∞                     ∞                    ∞                                        

8) For 0 < θ < Π/2, if x =  cos2nθ , y= sin2nθ , z= cos2nsin2nθ then

                                        n=0                     n=0                    n=0


(a)xyz = xz + y               (b)xyz = xy +z                (c)xyz = yz +x               (d) none of these                                     

(b) Here x = 1/(1-cos2θ) = cosec2θ,

y = sec2θ,

z = 1/(1-sin2θcos2θ)

⇒ z = 1/1-1/x.1/y = xy/xy-1

⇒ xyz = xy + z .

9) If a, b and c are distinct positive real numbers and a2 +b2 +c2 =1, then ab + bc +ca is

(a) less than 1               (b) equal to 1                (c) greater than 1                 (d) any real number.

(a) Since a and b are unequal , (a2+b2)/2>√a2b2(A.M. > G.M. for unequal numbers)

⇒ a2 +b> 2ab.

Similarly b2 +c2 > 2bc and c2 +a2 > 2ca. Hence 2(a2 +b2 +c2 ) > 2(ab + bc +ca)

⇒ ab +bc +ca < 1.

Hence (A) is the correct answer. Alternative solution: It is given that 1 = a2 + b2 + c2

= (a2+b2)/2 + (b2+c2)/2 + (c2+a2)/2 > ab+bc+ca

10) If the product of n positive numbers is unity , then their sum is

(a) a positive integer              (b) divisible by n               (c) equal to (n +1)/n                (d) never less than n.

(d) Let the numbers be a1; a2; a3 ;. . . ; an . Then a1. a2. a3 . . . . an =1.

Using A.M. ≥ G.M ,

we get 1/n(a1+a2+a3+….+an) ≥ (a1.a2.a3.  …..  .an)1/n

⇒ a1 + a2 + a3 +…. + an ≥ n

11) Suppose a, b, c are in A.P. and a2, b2, c2 are in G.P. If a < b < c and a+b+c = 3/2 , then the value of a is

(a) 1/2-1/√2               (b) 1/2√2                (c) 1/2+1/√2                 (d) none of these

(a) Since a + b + c = 3/2 , a + c = 1

⇒ b = 1/2
Also b4 = a2c2

⇒ b2 = ± ac.

⇒ac = -1/4 , (a < b < c)

Hence, a, c are the roots of the equation 4x2 – 4x – 1 = 0.

12) If the first and the (2n-1)th terms of an A.P., a G.P. and an H.P. are equal and their nth terms are a, b, c respectively , then

(a) a+c = 2b               (b) a + c = b                (c) ac –b2 = 0                 (d) none of these.

(c) Let α be the first and β be the (2n– 1)th terms.

Then α, a, β will be in A.P., α, b, β will be in G.P. α, c, β will be in H.P.

Hence a, b, c are respectively A. M. , G.M. and H.M. of α and β.

⇒ a = (α+β)/2 , b2 = αβ and c = 2αβ/(α+β).

Hence ac-b2 =0

13) If a, b, c and d are different positive numbers in H.P., then

(a) a+b > c+d               (b) a+c > b+d                (c) a+d > b+c                  (d) none of these.

(c) Since b is the H.M. of a and c, (a+c)/2  > b (A.M. > H.M.).

Again c is the H.M. of b and d, (b+d)/2 > c ( A.M. > H.M.).

Adding, we get (a+c)/2 + (b+d)/2 > b+c

⇒ a + d > b+ c.

14) Given p A.P’s, each of which consists of n terms. If their first terms are 1, 2, 3,……, p and common differences are 1, 3, 5, ….., 2p – 1 respectively , then sum of the terms of all the progressions is

(a) 1/2 [np(np+1)]               (b) 1/2 [n(p+1)]                (c) np(n+1)                  (d) none of these.

(a) The rth A. P. has first term r and common difference 2r-1.

 

Hence sum of its n terms = n/2[2r+(n-1)(2r-1)]

                                p

The required sum = n/2 [2r+(n-1)(2r-1)]

                               r=1
                               

        p                           p        p

= n  [r+n(n-1)]/2[2 r – 1]

      r=1                        r=1     r=1

 

=[np(p+1)]/2+[n(n+1)]/2[p(p+1)-p]

 

=np/2(np+1)      

15) The pth term of an A.P. is a and qth term is b. then the sum of it’s (p + q) terms is

(a) (p+q)/2[a+b+(a-b)/(p-q)]                                       (b) (p+q)/2[a-b-(a-b)/(p-q)]                

(c) (p+q)/2[a+b+(a+b)/(p+q)]                                     (d) none of these.

(a) Let x be the first term and d be the c . d of A.P.

a = x + (p – 1 ) d

b = x + ( q – 1)d

⇒ d =a-b/p-q    . . . . . (1)

so, x = a – (p-1)(a-b)/p-q

= (pa-qa-pa+pb+a-b)/(p-q) = (pb-qa+a-b)/(p-q)

Hence, Sp+q = (p+q)/2 [a+b+(a-b)/(p-q)

16) If (a2a3)/(a1a4) = 3(a2 – a3)/(a1 – a4) , then a1, a2, a3, a4 are in

(a) A.P               (b) G.P                (c) H.P                  (d) none of these.

(c) (a1+a4)/a1a4 = (a2+a3)/a2a3

⇒ 1/a4 + 1/a1 = 1/a2 + 1/a3

⇒1/a2 – 1/a1 = 1/a4 – 1/a3

 Also, 3(a2-a3)/a2a3 = (a1 – a4)/a1a4

⇒3(1/a3 – 1/a2) = 1/a4 – 1/a1

So, 1/a1 , 1/a2 ,1/a3 , 1/a4  are in A.P.

17) If ar > 0, r ∈ N and a1, a2, a3, ……, a2n are in A.P, then (a1+a2n)/(√a1+√a2) + (a2+a2n-1)/(√a2+√a3) + …… + (an+an+1)/(√an+√an+1) is equal to

(a) n-1               (b) n(a1+a2n)/(√a1+√an+1              (c) (n-1)/(√a1+√an+1)                  (d) none of these.

(b) a1 + a2n = a2 + a2n – 1 = ……… = an + an + 1 = k (say)

now given expression = k[(√a1 – √a2)/(a1 – a2)+….+(√an-√an+1)/(an – an+1)]

= -k/d(√an-√an+1)

= -k/d[(a1 – an+1)/(√a1+√an+1)]

= (a1 + a2n)(-nd)/-d(√a1+√an+1)

= n(a1+a2n)/(√a1+√an+1)

18) The greatest value of x2y3 is, where x > 0 and y > 0 are connected by the relation 3x + 4y =

(a) 3/16               (b) 1/16               (c) 3/14                  (d) none of these.

(a) We are given 3x + 4y = 5 ……(1)

Now, product x2y3 will be maximum when [(3/2)x]2 [(4/3)y]3 is maximum.

Above is product of 5 factors whose sum is 2.(3/2)x + 3.(4/3)y = 3x + 4y = 5 = constant Hence, the

product will be maximum when all the factors are equal

i.e.3x/2 = 4y/3 = (3x+4y)/(2+3) = 5/5 = 1 

⇒ x =2/3 , y = 3/4

∴ Maximum value of x2y3 = (2/3)2 (3/4)2

= 3/16

19) Let p, q, r ∈ R+ and 27 pqr ≥ (p + q + r)3 and 3p + 4q + 5r = 12 then p3 + q4 + r5 is equal to

(a) 3              (b) 6               (c) 2                  (d) none of these.

(a) 27 pqr ≥ ( p + q + r )3

⇒ (pqr)1/3 ≥ (p +q+r)/3

⇒ p = q = r

Also 3 p + 4q + 5r =12

⇒ p = q = r =1 .

20) Sum of infinite number of terms of GP is 20 and sum of their square is 100. The common ratio of GP is

(a) 5              (b) 3/5               (c) 8/5                  (d) 1/5

(b) Let = first term of and common ratio of 

Then  is ,  

Given

Also to

From 

From  we get 

⇒ 400(1-r)

⇒ 5r = 3

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