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 +b2 > 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 a = first term of and
Then is a ,
Given
Also to
From
From we get
⇒ 400(1-r)2
⇒ 5r = 3
