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

Question 1

If πi=1in(n+1)2, then πi=1(3i2)

A.

n(3n1)2

B.

n(3n+1)2

C.

n(3n + 2)

D.

n(3n+1)4

SOLUTION

Solution : A

πi=1 = 3πi=1i - 2 πi=11 = 3 n(n+1)2 - 2n =  n(3n1)2

Question 2

11.212.313.4 +.......+ ......... 1n.(n+1) equals

A.

1n(n+1)

B.

nn+1

C.

2nn+1

D.

2n(n+1)

SOLUTION

Solution : B

( 1112) + ( 1213) + ( 1314) + .........+ ( 1n1n+1)

= 1 -  1n+1nn+1

Question 3

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

A.

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

B.

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

C.

n2(n+2)

D.

n(n+1)(n+2)

SOLUTION

Solution : A

Tπn2 + n  ⇒ Sπ = Tπ =   n2n

n(n+1)(2n+1)6n(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?

A.

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

B.

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

C.

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

D. None of the above

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 1bc1ca1ab be consecutive terms of an A.P., then (bc)2,(ca)2,(ab)2 will be in ___.

A.

G.P

B.

A.P

C.

H.P

D.

None of these

SOLUTION

Solution : B

If (bc)2,(ca)2,(ab)2 are in A.P.

Then we have (ca)2(bc)2 = (ab)2(ca)2

(ba)(2cab) = (c - b)(2a - b - c)        ......(i)

Also if 1bc,1ca,1ab are in A.P.

Then 1ca1bc = 1ab1ca

b+a2c(ca)(bc) = c+b2a(ab)(ca)

(ab)(b+a2c) = (bc)(c+b2a)

(ba)(2cab) = (cb)(2abc)

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

A.

G.P

B.

A.P

C.

H.P

D.

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 = TqTpTrTqTqTrTpTq

= [a+(q1)d][a+(r1)d][a+(p1)d][a+(q1)d] = qrpq

Similarly, we can show that R =qrpq = rsqr

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.

A.P

B.

G.P

C.

H.P

D.

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.

A.P

B.

H.P

C.

G.P

D.

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

A.

515

B.

315

C.

110

D.

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+(51)(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

A.

rmn

B.

mnr+1

C.

mnr

D.

mnr1

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+(m1)d = 1n      ......(i)

a+(n1)d = 1m      ......(i)

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

Now, rth term of corresponding A.P.

= a+(r1)d = 1mn+(r1)1mn = 1+r1mn = 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 pthqth, and rth terms of a G.P., then (cb)p(ba)r(ac)q is equal to

A.

1

B.

apbqcr

C.

aqbrcp

D.

arbpcq

SOLUTION

Solution : A

a = ARp1, b = ARq1, c = ARr1

(cb)p(ba)r(ac)q = (ARp1ARq1)p(ARq1ARp1)r(ARp1ARr1)q

= R(rq)p+(qp)r+(pr)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

A.

34

B.

12

C.

13

D.

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

A.

1

B.

2

C.

3

D.

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

A.

3ad

B.

(a + b)(c + d)

C.

3ac

D.

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(bc)x2+b(ca)x+c(ab) = 0 be equal then a,b,c are in

A.

A.P

B.

G.P

C.

H.P

D.

None of these

SOLUTION

Solution : C

Since the roots of the quadratic equation in x are equal, we have (B24AC) = 0 

b2(ca)24ac(bc)(ab) = 0

b2(c22ca+a2)4ac(bab2ca+bc) = 0

b2(c2+2ca+a2)4ac{b(a+c)ac} = 0

b2(a+c)24ac{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.