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

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

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

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

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.

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).

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.

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.

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

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.

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.

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

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

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.

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

A.

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

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

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.