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## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.10

Multiple choice questions:

Question 1.

Euclid’s division lemma states that for positive integers a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy ………………….

(1) 1 < r < b

(2) 0 < r < b

(3) 0 < r < 6

(4) 0 < r < b

Ans.

(3) 0 < r < b

Question 2.

Using Euclid’s division lemma, if the cube of any positive integer is divided by 9 then the possible remainders are ………………….

(1) 0, 1, 8

(2) 1, 4, 8

(3) 0, 1, 3

(4) 1, 3, 5

Answer:

(1) 0, 1, 8

Hint: Let the +ve integer be 1, 2, 3, 4 …………

1^{3} = 1 when it is divided by 9 the remainder is 1.

2^{3} = 8 when it is divided by 9 the remainder is 8.

3^{3} = 27 when it is divided by 9 the remainder is 0.

4^{3} = 64 when it is divided by 9 the remainder is 1.

5^{3} = 125 when it is divided by 9 the remainder is 8.

The remainder 0, 1, 8 is repeated.

Question 3.

If the H.C.F of 65 and 117 is expressible in the form of 65m – 117 , then the value of m is

(1) 4

(2) 2

(3) 1

(4) 3

Answer:

(2) 2

Hint:

H.C.F. of 65 and 117

117 = 65 × 1 + 52

65 = 52 × 1 + 13

52 = 13 × 4 + 0

∴ 13 is the H.C.F. of 65 and 117.

65m – 117 = 65 × 2 – 117

130 – 117 = 13

∴ m = 2

Question 4.

The sum of the exponents of the prime factors in the prime factorization of 1729 is …………………….

(1) 1

(2) 2

(3) 3

(4) 4

Answer:

(3) 3

Hint: 1729 = 7 × 13 × 19

Sum of the exponents = 1 + 1 + 1

= 3

Question 5.

The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is

(1) 2025

(2) 5220

(3) 5025

(4) 2520

Answer:

(4) 2520

Hint:

L.C.M. = 2^{3} × 3^{2} × 5 × 7

= 8 × 9 × 5 × 7

= 2520

Question 6.

7^{4k} ≡ ______ (mod 100)

(1) 1

(2) 2

(3) 3

(4) 4

Answer:

(1) 1

Hint:

7^{4k} ≡______ (mod 100)

y^{4k} ≡ y^{4 × 1} = 1 (mod 100)

Question 7.

Given F_{1} = 1 , F_{2} = 3 and F_{n} = F_{n-1} + F_{n-2} then F_{5} is ………….

(1) 3

(2) 5

(3) 8

(4) 11

Answer:

(4) 11

Hint:

F_{n} = F_{n-1} + F_{n-2}

F_{3} = F_{2} + F_{1} = 3 + 1 = 4

F_{4} = F_{3} + F_{2} = 4 + 3 = 7

F_{5} = F_{4} + F_{3} = 7 + 4 = 11

Question 8.

The first term of an arithmetic progression is unity and the common difference is 4. Which of the following will be a term of this A.P

(1) 4551

(2) 10091

(3) 7881

(4) 13531

Answer:

(3) 7881

Hint:

t_{1} = 1

d = 4

t_{n} = a + (n – 1)d

= 1 + 4n – 4

4n – 3 = 4551

4n = 4554

n = will be a fraction

It is not possible.

4n – 3 = 10091

4n = 10091 + 3 = 10094

n = a fraction

4n – 3 = 7881

4n = 7881 + 3 = 7884

n = \(\frac{7884}{4}\), n is a whole number.

4n – 3 = 13531

4n = 13531 – 3 = 13534

n is a fraction.

∴ 7881 will be 1971st term of A.P.

Question 9.

If 6 times of 6^{th} term of an A.P is equal to 7 times the 7^{th} term, then the 13^{th} term of the A.P. is ………..

(1) 0

(2) 6

(3) 7

(4) 13

Answer:

(1) 0

Hint:

6 t_{6} = 7 t_{7}

6(a + 5d) = 7 (a + 6d) ⇒ 6a + 30d = 7a + 42d

30 d – 42 d = 7a – 6a ⇒ -12d = a

t_{13} = a + 12d (12d = -a)

= a – a = 0

Question 10.

An A.P consists of 31 terms. If its 16th term is m, then the sum of all the terms of this A.P. is

(1) 16 m

(2) 62 m

(3) 31 m

(4) \(\frac { 31 }{ 2 } \) m

Answer:

(3) 31 m

Hint:

t_{16} = m

S_{31} = \(\frac { 31 }{ 2 } \) (2a + 30d)

= \(\frac { 31 }{ 2 } \) (2(a + 15d))

(∵ t_{16} = a + 15d)

= 31(t_{16}) = 31m

Question 11.

In an A.P., the first term is 1 and the common difference is 4. How many terms of the A.P must be taken for their sum to be equal to 120?

(1) 6

(2) 7

(3) 8

(4) 9

Answer:

(3) 8

Here a = 1, d = 4, S_{n} = 120

S_{n} = \(\frac { n }{ 2 } \)[2a + (n – 1)d]

120 = \(\frac { n }{ 2 } \) [2 + (n – 1)4] = \(\frac { n }{ 2 } \) [2 + 4n – 4)]

= \(\frac { n }{ 2 } \) [4n – 2)] = \(\frac { n }{ 2 } \) × 2 (2n – 1)

120 = 2n^{2} – n

∴ 2n^{2} – n – 120 = 0 ⇒ 2n^{2} – 16n + 15n – 120 = 0

2n(n – 8) + 15 (n – 8) = 0 ⇒ (n – 8) (2n + 15) = 0

n = 8 or n = \(\frac { -15 }{ 2 } \) (omitted)

∴ n = 8

Question 12.

A = 2^{65} and B = 2^{64} + 2^{63} + 2^{62} …. + 2^{0} which of the following is true?

(1) B is 2^{64} more than A

(2) A and B are equal

(3) B is larger than A by 1

(4) A is larger than B by 1

Answer:

(4) A is larger than B by

A = 2^{65}

B = 2^{64+63} + 2^{62} + …….. + 2^{0}

= 2

= 1 + 2^{2} + 2^{2} + ……. + 2^{64}

a = 1, r = 2, n = 65 it is in G.P.

S_{65} = 1 (2^{65} – 1) = 2^{65} – 1

A = 2^{65} is larger than B

Question 13.

The next term of the sequence \(\frac { 3 }{ 16 } \),\(\frac { 1 }{ 8 } \),\(\frac { 1 }{ 12 } \),\(\frac { 1 }{ 18 } \) is ………..

(1) \(\frac { 1 }{ 24 } \)

(2) \(\frac { 1 }{ 27 } \)

(3) \(\frac { 2 }{ 3 } \)

(4) \(\frac { 1 }{ 81 } \)

Answer:

(2) \(\frac { 1 }{ 27 } \)

Hint:

\(\frac { 3 }{ 16 } \),\(\frac { 1 }{ 8 } \),\(\frac { 1 }{ 12 } \),\(\frac { 1 }{ 18 } \)

a = \(\frac { 3 }{ 16 } \), r = \(\frac { 1 }{ 8 } \) ÷ \(\frac { 3 }{ 16 } \) = \(\frac { 1 }{ 8 } \) × \(\frac { 16 }{ 3 } \) = \(\frac { 2 }{ 3 } \)

The next term is = \(\frac { 1 }{ 18 } \) × \(\frac { 2 }{ 3 } \) = \(\frac { 1 }{ 27 } \)

Question 14.

If the sequence t_{1},t_{2},t_{3} … are in A.P. then the sequence t_{6},t_{12},t_{18} … is

(1) a Geometric Progression

(2) an Arithmetic Progression

(3) neither an Arithmetic Progression nor a Geometric Progression

(4) a constant sequence

Answer:

(2) an Arithmetic Progression

Hint:

If t_{1}, t_{2}, t_{3}, … is 1, 2, 3, …

If t_{6} = 6, t_{12} = 12, t_{18} = 18 then 6, 12, 18 … is an arithmetic progression

Question 15.

The value of (1^{3} + 2^{3} + 3^{3} + ……. + 15^{3}) – (1 + 2 + 3 + …….. + 15) is …………….

(1) 14400

(2) 14200

(3) 14280

(4) 14520

Answer:

(3) 14280

Hint:

1202 – 120 = 120(120 – 1)

120 × 119 = 14280