# Free Objective Test 02 Practice Test - 11th and 12th

If p, q, r are in A.P. and are positive, the roots of the quadratic

equation p x2 + qx + r = 0 are all real for ___.

A.

| rp - 7| ≥ 4 3

B.

| pr - 7| < 4 3

C.

All p and r

D.

No p and r

#### SOLUTION

Solution : A

p,q,r  are positive and are in A.P.

∴ q =  p+r2              .........(i)

∵ The roots of p x2 + qx + r = 0 are real

q2 ≥ 4pr ⇒  [p+r2]2 ≥ 4pr  [using (i)]

p2r2 - 14pr ≥ 0

(rp)2 - 14 (rp) + 1 ≥ 0      ( ∵ p>0 and p ≠ 0)

(rp7)2  -  (43)2  ≥ 0

⇒| rp - 7| ≥ 4 3.

Sum of first n terms in the following series

cot13 +  cot17 +  cot113 +  cot121 + .........n terms.is given by

A.

tan1( nn+2)

B.

cot1( n+2n)

C.

tan1(n+1) -  tan1 1

D.

All of these

#### SOLUTION

Solution : D

Let S = 3 + 7 + 13 + 21 + ...... + Tπ  ⇒ Tπ  =  n2 + n + 1.

Let Trcot1( r2 + r + 1) =  tan1(r + 1) -  tan1r.

Put r = 1, 2,........, n and add, we get the required sum

tan1(n+1) -  tan11 =  tan1( nn+2) =cot1  (n+2n)

The sum of the series  12.2 +  22.3 +  32.4 + ........ to n terms is

A.

n3(n+1)3(2n+1)24

B.

n(n+1)(3n2+7n+2)12

C.

n(n+1)6[n(n+1) + (2n + 1)]

D.

n(n+1)12[6n(n+1) + 2(2n + 1)]

#### SOLUTION

Solution : B

Tπn2(n + 1) =  n3n2

Sπ = Tπn3 + n2 = [n(n+1)2]2 + n(n+1)(2n+1)6

n(n+1)2[n(n+1)22n+13] =  n(n+1)(3n2+7n+2)12

1 . The sum of the series 1 + 3x + 6x2 + 10 x3 + ....... ∞ will be

A.

1(1x)2

B.

11x

C.

1(1+x)2

D.

1(1x)3

#### SOLUTION

Solution : D

Let S = 1 + 3x + 6x2 + 10 x3 + ....... ∞

⇒ x.S = x + 3x2  + 6x3 + ..........∞

Subtracting S(1-x) = 1 + 2x + 3x2 + 4x3 + ........∞

⇒ x(1 - x)S = x + 2x2 + 3x3 + .........∞

Again subtracting,

⇒ S[(1 - x) - x(1 - x)] = 1 + x + x2 + x3 + ........... ∞

⇒ S[(1 - x)(1 - x)] =  11x ⇒ S = 1(1x)3

The sum of infinite terms of the following series

1 +  45 +  752 + 1053 + ...........∞ will be

A.

316

B.

358

C.

354

D.

3516

#### SOLUTION

Solution : D

Let the sum to infinity of the arithmetic-geometric series be

S = 1 + 4 .  15 + 7. 152 + 10. 153 +........

⇒ 15 S = 15 + 4.152 + 7.153 +...........

Subtracting (1 - 15)S = 1 + 3.15 + 3.152 + 3. 153 + .......

= 1 + 3(15152 + ..............)

⇒ 45 S =  1 + 3.15(1115) = 1 + 3474 ⇒ S = 3516.

Aliter :  Use direct formula S =  ab1r +  dbr(1r)2

Here a = 1, b = 1, d = 3, r =  15, therefore

S =  1115 +  3×1×15(1152)543516255415163516.

Aliter: Use S = [1 + r1r × diff. of A.P. ]11r

If a and b are two different positive real numbers, then which of the following relations is true

A.

2ab>(a+b)

B.

2ab<(a+b)

C.

2ab = (a+b)

D.

None of these

#### SOLUTION

Solution : B

We know that A > G > H

Where A is arithmetic mean, G is geometric mean and H is harmonic mean, then A > G

a+b2>ab or (a + b) > 2ab

If 1ba+1bc = 1a+1c, then a,b,c are in

A.

A.P

B.

G.P

C.

H.P

D.

In G.P and H.P both

#### SOLUTION

Solution : C

Since the reciprocals of a and c occur on RHS, let us first assume that a,b,c are in H.P.

So that 1a,1b,1c are in A.P.

1b1a = 1c1b = d,say

abab = d = bcbcab = abd and b - c =bcd

Now LHS = 1ab+1b+c = 1abd+1bcd

=1bd(1c1a) = 1bd(2d)2b = 1a+1c = RHS

a, b, c are in H.P. is verified.

If sixth term of a H.P. is 161 and its tenth term is 1105, then first term of that H.P. is

A.

128

B.

139

C.

16

D.

117

#### SOLUTION

Solution : C

T6 of H.P. = 161T6 of A.P. = 61

and T10 of H.P. = 1105T10 of A.P. = 105,

so, a + 5d = 61 ........(i) and a + 9d = 105   .....(ii)

From (i) and (ii), a= 6

Therefore first term of H.P. = 1a = 16.

If the 7th term of a H.P is 110 and the 12th term is 125, then the 20th term is

A.

137

B.

141

C.

145

D.

149

#### SOLUTION

Solution : D

Considering corresponding A.P.

a + 6d = 10 and a + 11d = 25 d= 3, a= -8

T20 = a + 19d = -8 + 57 = 49

Hence 20th term of the corresponding H.P. is 149

If a, b, c are three distinct positive real numbers which are in H.P., then 3a+2b2ab+3c+2b2cb is

A.

Greater than or equal to 10

B.

Less than or equal to 10

C.

Only equal to 10

D.

None of these

#### SOLUTION

Solution : D

we have 1a,1b,1c are in A.P. Let 1a = p - q, 1b = p and

1c = p + q, where p,q > 0 and p > q. Now, substitute these

values in 3a+2b2ab+3c+2b2cb then it reduces to

10+14q2p2q2 which is obviously greater than

10(as p > q > 0).

Trick: Put a = 1, b = 12, c = 13.

The expression has the value 3+1212+1+12312 = 83+12>0

The first term of a harmonic is 17 and the second term is 19. The 12th term is

A.

119

B.

129

C.

117

D.

127

#### SOLUTION

Solution : B

Here first term of A.P. be 7 and second be 9, then 12th term will be 7 + 11 ×2 = 29

Hence 12th term of the H.P. be 129.

If x, y, z are in G.P. and axby = cz, then

A.

logac = logba

B.

logba = logcb

C.

logcb = logac

D.

None of these

#### SOLUTION

Solution : B

x, y, z are in G.P., then y2 = x.z

Now axby = cz = m

xlogda = ylogdb = zlogdc = logdm

x = logam, y=logbm, z = logcm

Again as x, y, z are in G.P., so yx = zy

logbmlogam = logcmlogbm logba = logcb

If the ratio of the sum of first three terms and the sum of first six terms of a G.P. be 125 : 152, then the common ratio r is ___.

A.

35

B.

53

C.

23

D.

32

#### SOLUTION

Solution : A

Given that a+ar+ar2a+ar+ar2+ar3+ar4+ar5=125152.

1+r+r21+r+r2+r3+r4+r5=125152    i.e., 1+r+r2(1+r+r2)+r3(1+r+r2)=125152                           11+r3=125152                               r3=1521251                                         =27125                                 r=35

The terms of a G.P. are positive. If each term is equal to the sum of two terms that follow it, then the common ratio is ___.

A.

512

B.

152

C.

1

D.

25

#### SOLUTION

Solution : A

The given condition can be expressed as
Tn = Tn+1+Tn+2, where n1.
If a is the first term and r is the common ratio, then by the condition is
arn1=arn+arn+1.rn1=rn(1+r)
r2+r1=0
r=1±1+42=1±52

Since each term is positive, the common ratio of the GP is 512.

If the sum of two extreme numbers of an A.P. with four terms is 8 and product of remaining two middle term is 15, then greatest number of the series will be

A. 5
B. 7
C. 9
D. 11

#### SOLUTION

Solution : B

Let four numbers are a3d, ad, a+d, a+3d.
Now (a3d)+(a+3d)=8a=4 and (ad)(a+d)=15a2d2=15d=1 Thus required numbers are 1,3,5,7. Hence greatest number is 7.