# Free Objective Test 03 Practice Test - 11th and 12th

### Question 1

If π∑i=1i = n(n+1)2, then π∑i=1(3i−2) =

n(3n−1)2

n(3n+1)2

n(3n + 2)

n(3n+1)4

#### SOLUTION

Solution :A

π∑i=1 = 3π∑i=1i - 2 π∑i=11 = 3 n(n+1)2 - 2n = n(3n−1)2

### Question 2

11.2 + 12.3 + 13.4 +.......+ ......... 1n.(n+1) equals

1n(n+1)

nn+1

2nn+1

2n(n+1)

#### SOLUTION

Solution :B

( 11 - 12) + ( 12 - 13) + ( 13 - 14) + .........+ ( 1n - 1n+1)

= 1 - 1n+1 = nn+1

### Question 3

The sum of n terms of the series whose nth term is n(n+1) is equal to

n(n+1)(n+2)3

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

n2(n+2)

n(n+1)(n+2)

#### SOLUTION

Solution :A

Tπ = n2 + n ⇒ Sπ = ∑Tπ = ∑ n2 + ∑n

= n(n+1)(2n+1)6 + n(n+1)2

= n(n+1)6 {2n + 1 + 3} = n(n+1)(n+2)3

### Question 4

If a1x=b1y=c1z and a,b,c are in G.P., then which of the following are true?

x,y,z will be in A.P.

x,y,z will be in G.P.

x,y,z will be in H.P.

#### SOLUTION

Solution :A

Let a1x=b1y=c1z=k.

⇒a=kx,b=ky,c=kz.Since a,b,c are in G.P., b2=ac.

∴k2y=kx.kz=kx+z

∴2y=x+z

⇒x,y,z are in A.P.

### Question 5

If 1b−c, 1c−a, 1a−b be consecutive terms of an A.P., then (b−c)2,(c−a)2,(a−b)2 will be in ___.

G.P

A.P

H.P

None of these

#### SOLUTION

Solution :B

If (b−c)2,(c−a)2,(a−b)2 are in A.P.

Then we have (c−a)2−(b−c)2 = (a−b)2−(c−a)2

⇒(b−a)(2c−a−b) = (c - b)(2a - b - c) ......(i)

Also if 1b−c,1c−a,1a−b are in A.P.

Then 1c−a−1b−c = 1a−b−1c−a

⇒b+a−2c(c−a)(b−c) = c+b−2a(a−b)(c−a)

⇒(a−b)(b+a−2c) = (b−c)(c+b−2a)

⇒(b−a)(2c−a−b) = (c−b)(2a−b−c)

Which is true by virtue of (i).

### Question 6

If pth,qth,rth and sth terms of an A.P. be in G.P., then (p - q),(q - r),(r - s) will be in

G.P

A.P

H.P

None of these

#### SOLUTION

Solution :A

If a and d be the first term and common difference of the A.P.

Then Tp = a + (p - 1)d, Tq = a + (q - 1)d and

Tr = a+(r - 1)d.

If Tp,Tq,Tr are in G.P.

Then its common ratio R = TqTp = TrTq = Tq−TrTp−Tq

= [a+(q−1)d]−[a+(r−1)d][a+(p−1)d]−[a+(q−1)d] = q−rp−q

Similarly, we can show that R =q−rp−q = r−sq−r

Hence (p - q), (q - r),(r - s) be in G.P.

### Question 7

If the numbers be in G.P., then their logarithms will be in

A.P

G.P

H.P

None of these

#### SOLUTION

Solution :A

Let three number a,b and c in G.P., then b2 = ac

⇒2logdb = logda+logdc or logdb = logda+logdc2

Thus their logarithms are in A.P.

### Question 8

If logax,logbx,logcx be in H.P., then a,b,c are in

A.P

H.P

G.P

None of these

#### SOLUTION

Solution :C

Here log xlog a, log xlog b, log xlog c are in H.P.

⇒log alog x, log blog x, log clog x are in A.P.

⇒logxa, logxb, logxc are in A.P.

⇒a,b,c are in G.P.

Note: Students should remember this question as afact

### Question 9

The fifth of the H.P. 2,52,103,................. will be

515

315

110

10

#### SOLUTION

Solution :D

Series, 2,52,103,................. are in H.P. ⇒12,25,310,......... will be in A.P.

Now first term a = 12 and common difference d = −110

So, 5th term of the A.P. = 12+(5−1)(−110) = 110

Hence 5th term in H.P. is 10.

### Question 10

If the mth term of a H.P. be n and nth be m, then the rth term will be

rmn

mnr+1

mnr

mnr−1

#### SOLUTION

Solution :C

Given Tm = n, Tn = m for H.P. therefore for the corresponding A.P. mth term = 1n, nth term = 1m

Let a and d be the first term and common difference of this A.P., then

a+(m−1)d = 1n ......(i)

a+(n−1)d = 1m ......(i)

Solving these, we get a = 1mn, d = 1mn

Now, rth term of corresponding A.P.

= a+(r−1)d = 1mn+(r−1)1mn = 1+r−1mn = rmn

Therefore rth term of corresponding H.P. is mnr

Note : Students should remember this question as a fact.

### Question 11

If a, b, c are pth, qth, and rth terms of a G.P., then (cb)p(ba)r(ac)q is equal to

1

apbqcr

aqbrcp

arbpcq

#### SOLUTION

Solution :A

a = ARp−1, b = ARq−1, c = ARr−1

∴(cb)p(ba)r(ac)q = (ARp−1ARq−1)p(ARq−1ARp−1)r(ARp−1ARr−1)q

= R(r−q)p+(q−p)r+(p−r)q = R0 = 1

### Question 12

If the 5th term of a G.P. containing positive terms is 13 and 9th term is 16243, then the 4th term will be

34

12

13

25

#### SOLUTION

Solution :B

T5 = ar4 = 13 .......(i)

and T9 = ar8 = 16243 .......(ii)

Solving (i) and (ii), we get r = 23 and a = 2716

Now 4th term = ar3 = 3324.2333 = 12

### Question 13

The number which should be added to the number added to the numbers 2, 14, 62 so that the resulting numbers may be in G.P. is

1

2

3

4

#### SOLUTION

Solution :B

Suppose that the added number be x

then x + 2, x + 14, x + 62 be in G.P.

Therefore (x+14)2 = (x + 2)(x + 62)

⇒x2+196+28x = x2+64x+124

⇒36x = 72 ⇒x = 2

Trick : (a) Let 1 is added, then the numbers will be 3, 15, 63 which are obviously not in G.P.

(b) Let 2 is added, then the numbers will be 4, 16, 64 which are obviously in G.P.

### Question 14

If a, b, c, d are in H.P., then ab + bc + cd is equal to

3ad

(a + b)(c + d)

3ac

None of these

#### SOLUTION

Solution :A

Since a, b, c, d are in H.P., therefore b is the H.M. of a and c

i.e., b = 2acb+d, ∴(a+c)(b+d) = 2acb.2bdc

⇒ ab+ad+bc+cd = 4ad ⇒ ab+bc+cd = 3ad

Trick : Check for a =1, b = 12, c = 13, d = 14

### Question 15

If the roots of a(b−c)x2+b(c−a)x+c(a−b) = 0 be equal then a,b,c are in

A.P

G.P

H.P

None of these

#### SOLUTION

Solution :C

Since the roots of the quadratic equation in x are equal, we have (B2−4AC) = 0

⇒b2(c−a)2−4ac(b−c)(a−b) = 0

⇒b2(c2−2ca+a2)−4ac(ba−b2−ca+bc) = 0

⇒b2(c2+2ca+a2)−4ac{b(a+c)−ac} = 0

⇒b2(a+c)2−4ac{b(a+c)−ac} = 0

Which can be seen to be true, if

b = 2aca+c or b(a+c) = 2ac i.e., if a,b,c are in H.P.