**1) If a, b , c, d and p are distinct real numbers such that (a ^{2} + b^{2} + c^{2}) p^{2} – 2(ab + bc + cd ) p + (b^{2} + c^{2} + d^{2}) ≤ 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)]

⇒16r^{2} – 16r + 3 = 0

⇒(4r – 1) (4r – 3) = 0

⇒ r = 1/4 ⇒ a = 3.

**3) The sum of the first n terms of the series 1 ^{2}+2.2^{2}+3^{2}+2.4^{2}+5^{2}+2.6^{2}+ . . . is n(n+1)^{2}/2, when n is even. When n is odd, the sum is **

(a) **[n ^{2}(n+1)]/2** (b)

**[n(n+1)(2n+1)]/6**(c)

**[n(n+1)**(d)

^{2}]/2**[n**

^{2}(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 = [n^{2}(n-1)]/2.

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

**4) Let α, β be the roots of x ^{2} – x + p = 0 and γ, δ be the roots of x^{2} – 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 (r^{2} + r^{3}) = 4

⇒ r2 = 4 ⇒ r = – 2.

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

p = αβ = α^{2}r = -2,

q = γδ = α^{2}r^{5} = – 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 AR^{p-1}, AR^{q-1} and AR^{r-1}.

Obviously

AR^{p-1}×AR^{r-1} = A^{2}R^{p+r-2} = (AR^{q-1})^{2},

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

**6) If 4a ^{2} + 9b^{2} + 16c^{2} = 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 4a^{2} + 9b^{2} + 16c^{2} = 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(a ^{r}/b^{r-1})=klog(a^{n+1}/b^{n-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(a^{r}-logb^{r-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 a^{n+1}/b^{n-1}]

** ∞ ∞ ∞ **

**8) For 0 < θ < Π/2, if x = ∑ cos ^{2n}θ , y=∑ sin^{2n}θ , z=∑ cos^{2n}sin^{2n}θ 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-cos^{2}θ) = cosec^{2}θ,

y = sec^{2}θ,

z = 1/(1-sin^{2}θcos^{2}θ)

⇒ 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 a ^{2} +b^{2} +c^{2} =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 , (a^{2}+b^{2})/2>√a^{2}b^{2}(A.M. > G.M. for unequal numbers)

⇒ a^{2} +b^{2 }> 2ab.

Similarly b^{2} +c^{2} > 2bc and c^{2} +a^{2} > 2ca. Hence 2(a^{2} +b^{2} +c^{2} ) > 2(ab + bc +ca)

⇒ ab +bc +ca < 1.

Hence (A) is the correct answer. Alternative solution: It is given that 1 = a^{2} + b^{2} + c^{2}

= (a^{2}+b^{2})/2 + (b^{2}+c^{2})/2 + (c^{2}+a^{2})/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 a_{1}; a_{2}; a_{3} ;. . . ; a_{n} . Then a_{1}. a_{2}. a_{3} . . . . a_{n} =1.

Using A.M. ≥ G.M ,

we get 1/n(a_{1}+a_{2}+a_{3}+….+a_{n}) ≥ (a_{1}.a_{2}.a_{3}. ….. .a_{n})^{1/n}

⇒ a_{1} + a_{2} + a_{3} +…. + a_{n} ≥ n

**11) Suppose a, b, c are in A.P. and a ^{2}, b^{2}, c^{2} 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 b^{4} = a^{2}c^{2}

⇒ b^{2} = ± ac.

⇒ac = -1/4 , (a < b < c)

Hence, a, c are the roots of the equation 4x^{2} – 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 –b ^{2} = 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 , b^{2} = αβ and c = 2αβ/(α+β).

Hence ac-b^{2} =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 p ^{th} term of an A.P. is a and q^{th} 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, S_{p+q} = (p+q)/2 [a+b+(a-b)/(p-q)

**16) If (a _{2}a_{3})/(a_{1}a_{4}) = 3(a_{2 }– a_{3})/(a_{1 }– a_{4}) , then a_{1}, a_{2}, a_{3}, a_{4} are in**

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

(c) (a_{1}+a_{4})/a_{1}a_{4} = (a_{2}+a_{3})/a_{2}a_{3}

⇒ 1/a_{4} + 1/a_{1} = 1/a_{2} + 1/a_{3}

⇒1/a_{2} – 1/a_{1} = 1/a_{4} – 1/a_{3}

Also, 3(a_{2}-a_{3})/a_{2}a_{3} = (a_{1} – a_{4})/a_{1}a_{4}

⇒3(1/a_{3} – 1/a_{2}) = 1/a_{4} – 1/a_{1}

So, 1/a_{1} , 1/a_{2} ,1/a_{3} , 1/a_{4} are in A.P.

**17) If a _{r} > 0, r ∈ N and a_{1}, a_{2}, a_{3}, ……, a_{2n} are in A.P, then (a_{1}+a_{2n})/(√a_{1}+√a_{2}) + (a_{2}+a_{2n-1})/(√a_{2}+√a_{3}) + …… + (a_{n}+a_{n+1})/(√a_{n}+√a_{n+1}) is equal to **

(a) **n-1** (b) **n(a _{1}+a_{2n})/(√a_{1}+√a_{n+1}) ** (c)

**(n-1)/(√a**(d)

_{1}+√a_{n+1})**none of these.**

(b) a_{1} + a_{2n} = a_{2} + a_{2n – 1} = ……… = a_{n} + a_{n + 1} = k (say)

now given expression = k[(√a_{1} – √a_{2})/(a_{1} – a_{2})+….+(√a_{n}-√a_{n+1})/(a_{n} – a_{n+1})]

= -k/d(√a_{n}-√a_{n+1})

= -k/d[(a_{1} – a_{n+1})/(√a_{1}+√a_{n+1})]

= (a_{1} + a_{2n})(-nd)/-d(√a_{1}+√a_{n+1})

= n(a_{1}+a_{2n})/(√a_{1}+√a_{n+1})

**18) The greatest value of x ^{2}y^{3} 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 x^{2}y^{3} 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 x^{2}y^{3} = (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 p^{3} + q^{4} + r^{5} 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 G.P and

Then G.P. is a , ar, ar^{2}

Given S_{∞ }= 20

⇒ a/(1−r) = 20

⇒ a = 20(1−r)....(i)

Also a^{2}+a^{2}r^{2}+a^{2}r^{4}+... to ∞=100

⇒ a^{2}/(1- r^{2}) = 100

⇒ a^{2 }= 100(1−r)(1+r)....(ii)

From (i), a^{2} = 400(1- r)^{2}

From (ii), we get 100(1−r)(1+r)

⇒ 400(1-r)^{2 }

⇒ 1+r = 4−4r

⇒ 5r = 3

⇒ r = 3/5.