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

### Question 1

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 ___.

| rp - 7| ≥ 4 √3

| pr - 7| < 4 √3

All p and r

No p and r

#### SOLUTION

Solution :A

p,q,rare 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)]

⇒p2 + r2 - 14pr ≥ 0

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

(rp−7)2 - (4√3)2 ≥ 0

⇒| rp - 7| ≥ 4 √3.

### Question 2

Sum of first n terms in the following series

cot−13 + cot−17 + cot−113 + cot−121 + .........n terms.is given by

tan−1( nn+2)

cot−1( n+2n)

tan−1(n+1) - tan−1 1

All of these

#### SOLUTION

Solution :D

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

Let Tr = cot−1( r2 + r + 1) = tan−1(r + 1) - tan−1r.

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

tan−1(n+1) - tan−11 = tan−1( nn+2) =cot−1 (n+2n)

### Question 3

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

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

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

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

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

#### SOLUTION

Solution :B

Tπ = n2(n + 1) = n3 + n2

Sπ = ∑Tπ = ∑n3 + ∑n2 = [n(n+1)2]2 + n(n+1)(2n+1)6

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

### Question 4

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

1(1−x)2

11−x

1(1+x)2

1(1−x)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)] = 11−x ⇒ S = 1(1−x)3

### Question 5

The sum of infinite terms of the following series

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

316

358

354

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(15 + 152 + ..............)

⇒ 45 S = 1 + 3.15(11−15) = 1 + 34 = 74 ⇒ S = 3516.

Aliter :Use direct formula S∞ = ab1−r + dbr(1−r)2Here a = 1, b = 1, d = 3, r = 15, therefore

S∞ = 11−15 + 3×1×15(1−152) = 54 + 351625 = 54 + 1516 = 3516.

Aliter:Use S = [1 + r1−r × diff. of A.P. ]11−r

### Question 6

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

2√ab>(a+b)

2√ab<(a+b)

2√ab = (a+b)

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) > 2√ab

### Question 7

If 1b−a+1b−c = 1a+1c, then a,b,c are in

A.P

G.P

H.P

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.

⇒1b−1a = 1c−1b = d,say

⇒a−bab = d = b−cbc⇒a−b = abd and b - c =bcd

Now LHS = −1a−b+1b+c = −1abd+1bcd

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

∴ a, b, c are in H.P. is verified.

### Question 8

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

128

139

16

117

#### SOLUTION

Solution :C

T6 of H.P. = 161⇒T6 of A.P. = 61

and T10 of H.P. = 1105⇒T10 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.

### Question 9

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

137

141

145

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

### Question 10

If a, b, c are three distinct positive real numbers which are in H.P., then 3a+2b2a−b+3c+2b2c−b is

Greater than or equal to 10

Less than or equal to 10

Only equal to 10

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+2b2a−b+3c+2b2c−b then it reduces to

10+14q2p2−q2 which is obviously greater than

10(as p > q > 0).

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

The expression has the value 3+12−12+1+123−12 = 83+12>0

### Question 11

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

119

129

117

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.

### Question 12

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

logac = logba

logba = logcb

logcb = logac

None of these

#### SOLUTION

Solution :B

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

Now ax = by = 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

### Question 13

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 ___.

35

53

23

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=152125−1 =27125 ⇒r=35

### Question 14

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 ___.

√5−12

−1−√52

1

2√5

#### SOLUTION

Solution :A

The given condition can be expressed as

Tn = Tn+1+Tn+2, where n≥1.

If a is the first term and r is the common ratio, then by the condition is

arn−1=arn+arn+1.⇒rn−1=rn(1+r)

⇒r2+r−1=0

⇒r=−1±√1+42=−1±√52Since each term is positive, the common ratio of the GP is √5−12.

### Question 15

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

#### SOLUTION

Solution :B

Let four numbers are a−3d, a−d, a+d, a+3d.

Now (a−3d)+(a+3d)=8⇒a=4 and (a−d)(a+d)=15⇒a2−d2=15⇒d=1 Thus required numbers are 1,3,5,7. Hence greatest number is 7.