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

### Question 1

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

16

14

2714

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

### Question 2

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

2

3

4

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 r2−1(r−1) = 5a (r2−1)(r2−1) ⇒ r + 1 = 5 ⇒ r = 4

### Question 3

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

RS

SR

(RS)n

(SR)n

#### SOLUTION

Solution :D

S=a(1−rn)1−rP=anrn(n−1)2R=1a+1ar+1ar2+⋯⋯+1arn−1=1a(1−1rn)1−1r=1arn−1(rn−1r−1)p2=a2nrn(n−1)(SR)=a2nrn(n−1)

### Question 4

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

7

8

9

10

#### SOLUTION

Solution :A

(b, c, d) Given xn = xn+1 √2

∴ x1 = x2√2, x2 = x3√2, xn = xn+1√2

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

Hence xn = x1(√2)n−1

Area of Sn = x2n = x2n2n−1 < 1 ⇒ 2n−1 > x21 (x1 = 10)

∴ 2n−1 > 100

But 27 > 100, 28>100, etc.

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

### Question 5

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

Rs. 21555

Rs. 20475

Rs. 20500

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.

### Question 6

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

nn+1

2nn+1

2n(n+1)

2(n+1)n+2

#### SOLUTION

Solution :D

Tn=1[n(n+1)2]=2[1n−1n+1]

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

T1=2[11−12],T2=2[12−13],...........,

Tn+1=2[1n+1−1n+2]

Hence sum of (n+1) terms = n+1∑k=1Tk

⇒ Sn+1=2[1−1n+2] ⇒ Sn+1=2(n+1)n+2

### Question 7

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

1

2

32

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 = 18 + 216 + 332 + 464 + ................∞ ..........(ii)

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

Hence required product = 21 = 2.

### Question 8

The sum of 1 + 25 + 352 + 453 + .............upto n terms is

2516 - 4n+516×5n−1

34 - 2n+516×5n+1

37 - 3n+516×5n−1

12 - 5n+13×5n+2

#### SOLUTION

Solution :A

Given series, let Sn = 1 + 25 + 352 + 453 + .............+ n5n−1

15Sn = 15 + 252 + 353 + ..............+ n5n

Subtracting,

(1 - 15)Sn = 1 + 15 +

152 + 153 + ............+ upto n terms - n5n⇒ 45 Sn = 1−15n45 - n5n ⇒ Sn = 2516 - 4n+516×5n−1

### Question 9

The sum of the first n terms is 12 + 34 + 78 + 1516 + .......... is

2n - n - 1

1 - 2−n

n + 2−n - 1

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 - { 12 + 122 +............+ 12n}

= n - 12(1−12n1−12) = n - (1 - 12n) = n - 1 + 2−n.

Trick:Check for n = 1, 2 i.e., S1 = 12, S2 = 54 and(c) ⇒ S1 = 12 and S2 = 2 + 2−2 - 1 = 54.

### Question 10

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

terms is

1−(n+1)xn+nxn+1(1−x)2

1−xn1−x

xn+1

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 xn−1 ..........(i)

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

Subtracting (ii) from (i), we get

(1-x)Sn = 1 + x + x2 + x3 + ......... to n terms -n xn

= ( (1−xn)(1−x)) - nxn

⇒ Sn = (1−xn)−nxn(1−x)(1−x)2 = 1−(n+1)xn+nxn+1(1−x)2

### Question 11

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

ab = cd

ad = bc

bc = ad

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.

### Question 12

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

A2 = GH

H2 = AG

G = AH

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

### Question 13

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

2

pqp+q

p+qpq

None of these

#### SOLUTION

Solution :A

As given H = 2pqp+q

∴Hp+Hq = 2qp+q+2pp+q = 2(p+q)p+q = 2

### Question 14

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

16

37

17

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.

### Question 15

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

mn

√mn

mn

o

#### SOLUTION

Solution :B

Given that

ap+q=arp+q−1=m and ap−q=arp−q−1=n.⇒m×n=arp+q−1 x arp−q−1

=a2r2(p−1)

=(arp−1)2

⇒√mn=arp−1=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 (p−q)th terms are equidistant from the pth term.

∴ pth term (√mn) will be G.M. of (p+q)th and (p−q)th terms.