# Free Objective Test 01 Practice Test - 11th and 12th

If the 7th term of a harmonic progression is 8 and the 8th term is 7, then its 15th term is

A.

16

B.

14

C.

2714

D.

5615

#### SOLUTION

Solution : D

Obviously, 7th term of corresponding A.P is 18 and 8th

term will be 17. a + 6d = 18 and a+7d = 17

Solving these, we get d = 156 and a = 156

Therefore 15th term of this A.P.

= 156+14×156 = 1556

Hence the required 15th term of the H.P. is 5615

A series in G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of the terms occupying odd places, then the common ratio will be equal to

A.

2

B.

3

C.

4

D.

5

#### SOLUTION

Solution : C

Let there be 2n terms in the given G.P. with first term a and the common ratio r.

Then, a  r21(r1) = 5a (r21)(r21) ⇒ r + 1 = 5 ⇒ r = 4

If the sum of the n terms of G.P. is S product is P and sum of their inverse is R, than  P2 is equal to

A.

RS

B.

SR

C.

(RS)n

D.

(SR)n

#### SOLUTION

Solution : D

S=a(1rn)1rP=anrn(n1)2R=1a+1ar+1ar2++1arn1=1a(11rn)11r=1arn1(rn1r1)p2=a2nrn(n1)(SR)=a2nrn(n1)

Let S1, S2,....... be squares such that for each n≥1, the length

of a side of Sn equals the length of a diagonal of Sn+1. If the

length of a  side of S1 is 10 cm, then for which of the following

values of n is the area of Sn greater then 1sq cm

A.

7

B.

8

C.

9

D.

10

#### SOLUTION

Solution : A

(b, c, d) Given xn = xn+1 2

∴ x1x22x2x32xnxn+12

On multiplying x1xn+1 (2)n   ⇒ xn+1 =   x1(2)n

Hence  xn =   x1(2)n1

Area of Sn = x2nx2n2n1 < 1  ⇒  2n1x21   (x1 = 10)

∴ 2n1 > 100

But 27 > 100, 28>100, etc.

∴ n - 1 = 7, 8, 9.......  ⇒ n = 8, 9, 10.........

Harikiran purchased a house in Rs. 15000 and paid Rs. 5000 at once. Rest money he promised to pay in annual instalment of Rs. 1000 with 10% per annum interest. How much money is to be paid by him

A.

Rs. 21555

B.

Rs. 20475

C.

Rs. 20500

D.

Rs. 20700

#### SOLUTION

Solution : C

It will take 10 years for harikiran to pay off Rs. 10000 in 10 yearly

installments.

∵ He pays 10% annual interest on interest on remaining amount

∴ Money given in first year

= 1000 +  10000×10100 = Rs. 2000

Money given in second year = 1000 + interest of (10000 - 1000) with interest rate 10% per annum

= 1000 +  9000×10100 = Rs. 1900

Money paid  in third year = Rs. 1800 etc.

So money given by Jairam in 10 years will be Rs. 2000, Rs.1900, Rs. 1800, Rs. 1700,......,

which is in arithmetic progression,

whose first term a = 2000 and d = -100

Total money given in 10 years = sum of 10 terms of arithmetic

progression

102[2(2000) + (10 - 1)(-100)] = Rs. 15500

Therefore, total money given by harikiran

= 5000 + 15500 = Rs. 20500.

The sum of (n+1) terms of 11+11+2+11+2+3+.......... is

A.

nn+1

B.

2nn+1

C.

2n(n+1)

D.

2(n+1)n+2

#### SOLUTION

Solution : D

Tn=1[n(n+1)2]=2[1n1n+1]

put n=1,2,3,.........,(n+1)

T1=2[1112],T2=2[1213],...........,

Tn+1=2[1n+11n+2]

Hence sum of (n+1) terms = n+1k=1Tk

Sn+1=2[11n+2]    Sn+1=2(n+1)n+2

214. 418. 8116. 16132................  is equal to

A.

1

B.

2

C.

32

D.

52

#### SOLUTION

Solution : B

214. 418. 8116. 16132 .........∞

214+28+316+........2S, where S is given by

S =  14 + 2 18 + 3 116 +4 132 + ............∞                              .........(i)

⇒  12 S =  18216 + 332 + 464 + ................∞     ..........(ii)

Subtracting (ii) from (i), we get S = 1.

Hence required product =  21 = 2.

The sum of 1 +  25352453 + .............upto n terms is

A.

2516 -   4n+516×5n1

B.

34 -   2n+516×5n+1

C.

37 -   3n+516×5n1

D.

12 -   5n+13×5n+2

#### SOLUTION

Solution : A

Given series, let Sn = 1 +  25352453 + .............+  n5n1

15Sn15252 +  353 + ..............+  n5n

Subtracting,

(1 - 15)Sn = 1 +  15 +  152 +  153 + ............+ upto n terms -  n5n

⇒  45 Sn115n45 -  n5n ⇒ Sn =   2516 -   4n+516×5n1

The sum of the first n terms is  1234781516 + .......... is

A.

2n - n - 1

B.

1 -  2n

C.

n +  2n - 1

D.

2n - 1

#### SOLUTION

Solution : C

The sum of the first n terms is

Sn = (1 - 12) + (1 - 122) + (1 - 123) + (1 - 124  + ........... + (1 - 12n)

= n - { 12122 +............+  12n}

= n -  12(112n112) = n - (1 - 12n) = n - 1 +  2n.

Trick: Check for n = 1, 2 i.e., S112S254 and

(c)  ⇒ S1 =   12 and S2 = 2 +  22  - 1 =  54.

The sum of the series 1 + 2x + 3 x2 + 4  x3 + ..........upto n

terms is

A.

1(n+1)xn+nxn+1(1x)2

B.

1xn1x

C.

xn+1

D.

None of these

#### SOLUTION

Solution : A

Let Sn be the sum of the given series to n terms, then

Sn = 1 + 2x + 3 x2 + 4 x3 + ....... + n xn1                         ..........(i)

xSn =       x + 2 x2 + 3 x2 + ...........+ n xn                          ..........(ii)

Subtracting (ii) from (i), we get

(1-x)Sn = 1 + x +  x2x3 + ......... to n terms -n xn

= ( (1xn)(1x)) - nxn

Sn(1xn)nxn(1x)(1x)21(n+1)xn+nxn+1(1x)2

If a,b,c be in A.P. and b,c,d be in H.P., then

A.

ab = cd

B.

C.

D.

abcd = 1

#### SOLUTION

Solution : C

Given that a,b,c in A.P. and b,c,d in H.P.

So, 2b = a + c and c = 2bdb+d

c(b + d) = 2bd = (a + c)d bc = ad.

If the aritmetic, geometric and harmonic menas between two positive real numbers be A, G and H, then

A.

A2 = GH

B.

H2 = AG

C.

G = AH

D.

G2 = AH

#### SOLUTION

Solution : D

Let A = a+b2, G =ab and H = 2aba+b.

Then G2 = ab            ..................(i)

and AH = (a+b2).2aba+b = ab    ................(ii)

From (i) and (ii), we have G2 = AH

If H is the harmonic mean between p and q, then the value of Hp+Hq is

A.

2

B.

pqp+q

C.

p+qpq

D.

None of these

#### SOLUTION

Solution : A

As given H = 2pqp+q

Hp+Hq = 2qp+q+2pp+q = 2(p+q)p+q = 2

The 4 term of a H.P is 35 and 8th term is 13, then its 6th term is

A.

16

B.

37

C.

17

D.

35

#### SOLUTION

Solution : B

a + 3d = 53 and a + 7d = 3

Solving a = 23, d = 13

6th term of A.P = a + 5d = 23+53 = 73

6th term of H.P. = 37.

If (p+q)th term and (pq)th term of a G.P. be m and n, then the pth term will be ___.

A.

mn

B.

mn

C.

mn

D.

o

#### SOLUTION

Solution : B

Given that
ap+q=arp+q1=m and apq=arpq1=n.

m×n=arp+q1arpq1

=a2r2(p1)

=(arp1)2

mn=arp1=ap

Thus, the pth term of the GP is mn.

Aliter: Each term in a G.P. is the geometric mean of the terms equidistant from it. Here, (p+q)th and (pq)th terms are equidistant from the pth term.
pth term (mn) will be G.M. of (p+q)th and (pq)th terms.