Samacheer Kalvi Class 10 Maths Solutions Chapter 2 Numbers and Sequences Exercise 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
(a) \( 1 < r < b \)
(b) \( 0 \le r < b \)
(c) \( 0 < r < b \)
Answer: (b) \( 0 \le r < b \)
In simple words: Euclid's division lemma is a fundamental rule in number theory. It shows that when you divide one positive integer by another, the remainder you get must be zero or a positive number, but it must always be smaller than the number you divided by.

🎯 Exam Tip: Remember the conditions for the remainder 'r' in Euclid's Division Lemma: \( 0 \le r < b \). This is a core concept that often appears in basic number theory questions.

Question 2. Using Euclid's division lemma, if the cube of any positive integer is divided by 9 then the possible remainders are
(a) 0, 1, 8
(b) 1, 4, 8
(c) 0, 1, 3
(d) 1, 3, 5
Answer: (a) 0, 1, 8
Hint: Let the positive 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.
In simple words: When you cube any positive whole number and then divide that cube by 9, the only remainders you will ever get are 0, 1, or 8. This pattern continues no matter which positive integer you start with.

🎯 Exam Tip: To find possible remainders, test the first few integers and look for a repeating pattern. This helps confirm the general rule without needing complex proofs.

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
(a) 4
(b) 2
(c) 1
Answer: (b) 2
Hint:
H.C.F. of 65 and 117
\( 117 = 65 \times 1 + 52 \)
\( 65 = 52 \times 1 + 13 \)
\( 52 = 13 \times 4 + 0 \)
\( \implies \) 13 is the H.C.F. of 65 and 117.
Given: \( 65m - 117 = \text{H.C.F.} \)
So, \( 65m - 117 = 13 \)
\( 65m = 13 + 117 \)
\( 65m = 130 \)
\( m = \frac{130}{65} \)
\( \implies m = 2 \)
In simple words: First, find the highest common factor (HCF) of 65 and 117 using Euclid's method. Once you have the HCF, set it equal to the given expression, which is 65m - 117, and solve for 'm'. This problem links the concept of HCF with simple algebraic equations.

🎯 Exam Tip: Always use the Euclidean algorithm to find HCF as it's efficient. Make sure to clearly state each step when equating the HCF to the given expression.

Question 4. The sum of the exponents of the prime factors in the prime factorization of 1729 is
(a) 1
(b) 2
(c) 3
Answer: (c) 3
Hint: First, find the prime factorization of 1729.
\( 1729 = 7 \times 13 \times 19 \)
The prime factors are 7, 13, and 19. Each of these factors has an exponent of 1.
Sum of the exponents \( = 1 + 1 + 1 \)
\( = 3 \)
In simple words: To solve this, first break down the number 1729 into its prime factors. Each prime factor will have a power (exponent) of 1. Then, just add these powers together to get the total sum.

🎯 Exam Tip: When finding prime factors, start with small prime numbers like 2, 3, 5, 7, etc., and divide until you are left with only prime numbers. Remember that if a prime factor appears once, its exponent is 1.

Question 5. The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is
(a) 5220
(b) 5025
(c) 2520
Answer: (c) 2520
Hint:
To find the least number divisible by all numbers from 1 to 10, calculate the Least Common Multiple (LCM) of these numbers.

L.C.M. \( = 2 \times 2 \times 3 \times 5 \times 7 \times 2 \times 3 = 2^3 \times 3^2 \times 5 \times 7 \)
\( = 8 \times 9 \times 5 \times 7 \)
\( = 2520 \)
In simple words: To find the smallest number that can be perfectly divided by every number from 1 to 10, you need to calculate their Least Common Multiple (LCM). This means finding the smallest number that is a multiple of all those numbers.

🎯 Exam Tip: When calculating LCM for a range of numbers, use the prime factorization method. Ensure all prime factors with their highest powers are included in the multiplication to get the correct LCM.

Question 6. \( 7^{4k} \equiv \_ \pmod{100} \)
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (a) 1
Hint:
We want to find \( 7^{4k} \pmod{100} \).
Let's look at the powers of 7 modulo 100:
\( 7^1 = 7 \pmod{100} \)
\( 7^2 = 49 \pmod{100} \)
\( 7^3 = 343 \equiv 43 \pmod{100} \)
\( 7^4 = 7 \times 43 = 301 \equiv 1 \pmod{100} \)
Since \( 7^4 \equiv 1 \pmod{100} \), then for any positive integer k:
\( 7^{4k} = (7^4)^k \equiv 1^k \pmod{100} \)
\( \implies 7^{4k} \equiv 1 \pmod{100} \)
In simple words: This question asks for the remainder when \( 7^{4k} \) is divided by 100. By checking the first few powers of 7, we find that \( 7^4 \) leaves a remainder of 1 when divided by 100. Because of this, any power of \( 7^4 \) will also leave a remainder of 1.

🎯 Exam Tip: For modular arithmetic problems involving powers, calculate the first few powers of the base modulo the given number until you find a remainder of 1. This cycle length can then be used to simplify higher powers.

Question 7. Given \( F_1 = 1, F_2 = 3 \) and \( F_n = F_{n-1} + F_{n-2} \) then \( F_5 \) is
(a) 3
(b) 5
(c) 8
(d) 11
Answer: (d) 11
Hint:
The formula for the sequence is \( F_n = F_{n-1} + F_{n-2} \).
We are given:
\( F_1 = 1 \)
\( F_2 = 3 \)
Now, let's find the next terms:
\( F_3 = F_2 + F_1 = 3 + 1 = 4 \)
\( F_4 = F_3 + F_2 = 4 + 3 = 7 \)
\( F_5 = F_4 + F_3 = 7 + 4 = 11 \)
In simple words: To find \( F_5 \), you need to follow the pattern where each number in the sequence is the sum of the two numbers before it. Start with \( F_1 \) and \( F_2 \) and keep adding to find \( F_3, F_4 \), and finally \( F_5 \).

🎯 Exam Tip: These types of sequence questions (like Fibonacci sequences) require careful step-by-step calculation. Write down each term to avoid mistakes, especially when dealing with slightly different starting values.

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
(a) 4551
(b) 10091
(c) 7881
(d) 13531
Answer: (c) 7881
Hint:
Given for the Arithmetic Progression (A.P.):
First term, \( t_1 = a = 1 \)
Common difference, \( d = 4 \)
The formula for the \( n^{th} \) term of an A.P. is \( t_n = a + (n - 1)d \).
Substitute the values: \( t_n = 1 + (n - 1)4 \)
\( t_n = 1 + 4n - 4 \)
\( t_n = 4n - 3 \)
Now, we check which of the given options can be a term by setting \( t_n \) equal to each option and solving for \( n \). If \( n \) is a whole number (positive integer), then the option is a term of the A.P.

For (a) 4551:
\( 4n - 3 = 4551 \)
\( 4n = 4554 \)
\( n = \frac{4554}{4} = 1138.5 \)
Since \( n \) is a fraction, 4551 is not a term.

For (b) 10091:
\( 4n - 3 = 10091 \)
\( 4n = 10094 \)
\( n = \frac{10094}{4} = 2523.5 \)
Since \( n \) is a fraction, 10091 is not a term.

For (c) 7881:
\( 4n - 3 = 7881 \)
\( 4n = 7884 \)
\( n = \frac{7884}{4} = 1971 \)
Since \( n \) is a whole number, 7881 is a term of the A.P. It is the \( 1971^{st} \) term.

For (d) 13531:
\( 4n - 3 = 13531 \)
\( 4n = 13534 \)
\( n = \frac{13534}{4} = 3383.5 \)
Since \( n \) is a fraction, 13531 is not a term.
In simple words: An arithmetic progression means numbers increase by the same amount each time. Here, the first number is 1, and it increases by 4 each time. To check if a number is part of this sequence, use the formula \( t_n = a + (n-1)d \). If solving for 'n' gives a whole number, then the option is a term in the sequence.

🎯 Exam Tip: When checking if a number is a term of an A.P., remember that the term number 'n' must always be a positive integer. If 'n' turns out to be a fraction, the number is not part of that specific 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
(a) 0
(b) 6
(c) 7
(d) 13
Answer: (a) 0
Hint:
Let 'a' be the first term and 'd' be the common difference of the A.P.
The \( n^{th} \) term is given by \( t_n = a + (n-1)d \).
So, \( t_6 = a + (6-1)d = a + 5d \)
And \( t_7 = a + (7-1)d = a + 6d \)
Given: \( 6 \times t_6 = 7 \times t_7 \)
\( 6(a + 5d) = 7(a + 6d) \)
\( 6a + 30d = 7a + 42d \)
Now, rearrange the terms to solve for 'a' in terms of 'd':
\( 30d - 42d = 7a - 6a \)
\( -12d = a \)
Now we need to find the \( 13^{th} \) term, \( t_{13} \):
\( t_{13} = a + (13-1)d = a + 12d \)
Substitute \( a = -12d \) into the expression for \( t_{13} \):
\( t_{13} = (-12d) + 12d \)
\( t_{13} = 0 \)
In simple words: For an arithmetic progression, if six times the sixth term is equal to seven times the seventh term, you can use the formula for the nth term to set up an equation. Solving this equation will show you a relationship between the first term 'a' and the common difference 'd', which you can then use to find the thirteenth term.

🎯 Exam Tip: Remember the general formula for the \( n^{th} \) term of an A.P., \( t_n = a + (n-1)d \). Algebraic manipulation is key here; be careful with signs when rearranging the equation to solve for 'a' or 'd'.

Question 10. An A.P consists of 31 terms. If its \( 16^{th} \) term is m, then the sum of all the terms of this A.P. is
(a) 16 m
(b) 62 m
(c) 31 m
(d) \( \frac { 31 }{ 2 } m \)
Answer: (c) 31 m
Hint:
Let 'a' be the first term and 'd' be the common difference.
The \( n^{th} \) term of an A.P. is \( t_n = a + (n-1)d \).
The sum of 'n' terms of an A.P. is \( S_n = \frac{n}{2} [2a + (n-1)d] \).

Given: The total number of terms is \( n = 31 \).
The \( 16^{th} \) term is \( m \). So, \( t_{16} = m \).
Using the \( n^{th} \) term formula: \( t_{16} = a + (16-1)d \)
\( m = a + 15d \)

Now, let's find the sum of all 31 terms, \( S_{31} \):
\( S_{31} = \frac{31}{2} [2a + (31-1)d] \)
\( S_{31} = \frac{31}{2} [2a + 30d] \)
Factor out 2 from the bracket:
\( S_{31} = \frac{31}{2} \times 2 [a + 15d] \)
\( S_{31} = 31 [a + 15d] \)
We know that \( a + 15d = m \). Substitute this into the sum:
\( S_{31} = 31m \)
In simple words: If you have an arithmetic progression with an odd number of terms, the middle term is very special. Here, with 31 terms, the 16th term is exactly in the middle. The sum of all terms is simply the total number of terms multiplied by this middle term.

🎯 Exam Tip: For an A.P. with an odd number of terms, the sum is always equal to the number of terms multiplied by the middle term. This shortcut can save time in calculations.

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?
(a) 6
(b) 7
(c) 8
(d) 9
Answer: (c) 8
Hint:
Given for the Arithmetic Progression (A.P.):
First term, \( a = 1 \)
Common difference, \( d = 4 \)
Sum of 'n' terms, \( S_n = 120 \)
The formula for the sum of 'n' terms of an A.P. is \( S_n = \frac{n}{2} [2a + (n-1)d] \).
Substitute the given values into the formula:
\( 120 = \frac{n}{2} [2(1) + (n-1)4] \)
\( 120 = \frac{n}{2} [2 + 4n - 4] \)
\( 120 = \frac{n}{2} [4n - 2] \)
Multiply both sides by 2:
\( 240 = n(4n - 2) \)
\( 240 = 4n^2 - 2n \)
Rearrange into a quadratic equation:
\( 4n^2 - 2n - 240 = 0 \)
Divide the entire equation by 2 to simplify:
\( 2n^2 - n - 120 = 0 \)
Now, solve the quadratic equation using factorization:
\( 2n^2 - 16n + 15n - 120 = 0 \)
Factor by grouping:
\( 2n(n - 8) + 15(n - 8) = 0 \)
\( (n - 8)(2n + 15) = 0 \)
This gives two possible values for 'n':
\( n - 8 = 0 \implies n = 8 \)
\( 2n + 15 = 0 \implies n = -\frac{15}{2} \)
Since the number of terms 'n' cannot be negative or a fraction, we take \( n = 8 \).
In simple words: This question asks how many terms from an arithmetic progression (starting at 1 and increasing by 4 each time) are needed to make their total sum 120. You use the sum formula for an A.P., set up a quadratic equation, and solve for 'n'. Only a positive whole number for 'n' makes sense.

🎯 Exam Tip: When solving for 'n' (number of terms) in A.P. or G.P. problems, always remember that 'n' must be a positive integer. Reject any negative or fractional solutions for 'n'.

Question 12. A = \( 2^{65} \) and B = \( 2^{64} + 2^{63} + 2^{62} + \ldots + 2^0 \) which of the following is true?
(a) B is \( 2^{64} \) more than A
(b) A and B are equal
(c) B is larger than A by 1
(d) A is larger than B by 1
Answer: (d) A is larger than B by 1
Hint:
Given:
\( A = 2^{65} \)
\( B = 2^{64} + 2^{63} + 2^{62} + \ldots + 2^0 \)
The expression for B is the sum of a Geometric Progression (G.P.).
Here, the first term \( a = 2^0 = 1 \).
The common ratio \( r = 2 \).
The number of terms \( n = 64 - 0 + 1 = 65 \).
The formula for the sum of a G.P. is \( S_n = \frac{a(r^n - 1)}{r - 1} \).
Substitute the values into the sum formula for B:
\( B = \frac{1(2^{65} - 1)}{2 - 1} \)
\( B = \frac{2^{65} - 1}{1} \)
\( B = 2^{65} - 1 \)
Now, compare A and B:
\( A = 2^{65} \)
\( B = 2^{65} - 1 \)
This means \( A - B = 2^{65} - (2^{65} - 1) = 2^{65} - 2^{65} + 1 = 1 \).
So, \( A = B + 1 \).
Therefore, A is larger than B by 1.
In simple words: The value B is a sum of powers of 2, starting from \( 2^0 \) up to \( 2^{64} \). This is a known sum for a geometric progression, which simplifies to \( 2^{65} - 1 \). Since A is \( 2^{65} \), it means A is exactly one more than B.

🎯 Exam Tip: Recognize sums of powers of 2 as geometric progressions. The sum \( 2^0 + 2^1 + \ldots + 2^{n-1} \) is always equal to \( 2^n - 1 \). This identity is very useful for quickly solving such problems.

Question 13. The next term of the sequence \( \frac { 3 }{ 16 }, \frac { 1 }{ 8 }, \frac { 1 }{ 12 }, \frac { 1 }{ 18 } \)
(a) \( \frac { 1 }{ 24 } \)
(b) \( \frac { 1 }{ 27 } \)
(c) \( \frac { 2 }{ 3 } \)
(d) \( \frac { 1 }{ 81 } \)
Answer: (b) \( \frac { 1 }{ 27 } \)
Hint:
Let's analyze the given sequence: \( \frac { 3 }{ 16 }, \frac { 1 }{ 8 }, \frac { 1 }{ 12 }, \frac { 1 }{ 18 } \)
This is a Geometric Progression (G.P.).
First term, \( a = \frac{3}{16} \).
To find the common ratio (r), divide the second term by the first term:
\( r = \frac { 1 }{ 8 } \div \frac { 3 }{ 16 } \)
\( r = \frac { 1 }{ 8 } \times \frac { 16 }{ 3 } \)
\( r = \frac { 16 }{ 24 } = \frac { 2 }{ 3 } \)
Let's check this ratio with the next terms:
\( \frac{1}{8} \times \frac{2}{3} = \frac{2}{24} = \frac{1}{12} \) (Matches the third term)
\( \frac{1}{12} \times \frac{2}{3} = \frac{2}{36} = \frac{1}{18} \) (Matches the fourth term)
The common ratio is indeed \( r = \frac{2}{3} \).
Now, to find the next term (the fifth term), multiply the fourth term by the common ratio:
Next term \( = \frac { 1 }{ 18 } \times \frac { 2 }{ 3 } \)
Next term \( = \frac { 2 }{ 54 } \)
Next term \( = \frac { 1 }{ 27 } \)
In simple words: Look for a pattern in the sequence by seeing how each term changes to the next. In this case, each term is multiplied by the same fraction to get the next term. Once you find this common fraction (the common ratio), multiply the last given term by it to find the next one.

🎯 Exam Tip: When dealing with sequences involving fractions, check if it's an Arithmetic Progression (A.P.) or a Geometric Progression (G.P.). For G.P., find the common ratio by dividing a term by its preceding term.

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
(a) a Geometric Progression
(b) an Arithmetic Progression
(c) neither an Arithmetic Progression nor a Geometric Progression
(d) a constant sequence
Answer: (b) an Arithmetic Progression
Hint:
Let the original sequence \( t_1, t_2, t_3, \ldots \) be an Arithmetic Progression (A.P.) with first term 'a' and common difference 'd'.
So, \( t_n = a + (n-1)d \).

The new sequence is formed by terms \( t_6, t_{12}, t_{18}, \ldots \)
Let's write out these terms:
\( t_6 = a + (6-1)d = a + 5d \)
\( t_{12} = a + (12-1)d = a + 11d \)
\( t_{18} = a + (18-1)d = a + 17d \)

To check if this new sequence is an A.P., we need to see if the difference between consecutive terms is constant.
Difference between \( t_{12} \) and \( t_6 \):
\( t_{12} - t_6 = (a + 11d) - (a + 5d) = a + 11d - a - 5d = 6d \)

Difference between \( t_{18} \) and \( t_{12} \):
\( t_{18} - t_{12} = (a + 17d) - (a + 11d) = a + 17d - a - 11d = 6d \)

Since the common difference between consecutive terms of the new sequence is constant (6d), the sequence \( t_6, t_{12}, t_{18}, \ldots \) is also an Arithmetic Progression.
The first term of this new A.P. is \( a + 5d \), and its common difference is \( 6d \).
Example: If the original A.P. is 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, ...
Then \( t_6 = 6 \)
\( t_{12} = 12 \)
\( t_{18} = 18 \)
The sequence 6, 12, 18, ... is an arithmetic progression with a common difference of 6.
In simple words: If you pick terms from an arithmetic progression at regular intervals (like every sixth term), those chosen terms will also form an arithmetic progression. The common difference of this new sequence will be a multiple of the original common difference.

🎯 Exam Tip: A key property of arithmetic progressions is that any subsequence formed by taking terms at equal intervals will also be an A.P. Use a simple example to quickly verify this property if you are unsure.

Question 15. The value of \( (1^3 + 2^3 + 3^3 + \ldots + 15^3) - (1 + 2 + 3 + \ldots + 15) \) is
(a) 14400
(b) 14200
(c) 14280
(d) 14520
Answer: (c) 14280
Hint:
We need to find the value of \( (1^3 + 2^3 + 3^3 + \ldots + 15^3) - (1 + 2 + 3 + \ldots + 15) \).
We use the standard formulas for the sum of the first 'n' natural numbers and the sum of the cubes of the first 'n' natural numbers.
Let \( n = 15 \).

Formula for the sum of the first 'n' natural numbers (\( \Sigma n \)):
\( 1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2} \)
For \( n=15 \):
\( 1 + 2 + 3 + \ldots + 15 = \frac{15(15+1)}{2} = \frac{15 \times 16}{2} = 15 \times 8 = 120 \)

Formula for the sum of the cubes of the first 'n' natural numbers (\( \Sigma n^3 \)):
\( 1^3 + 2^3 + 3^3 + \ldots + n^3 = \left( \frac{n(n+1)}{2} \right)^2 \)
For \( n=15 \):
\( 1^3 + 2^3 + 3^3 + \ldots + 15^3 = \left( \frac{15(15+1)}{2} \right)^2 = \left( \frac{15 \times 16}{2} \right)^2 \)
\( = (15 \times 8)^2 = (120)^2 = 14400 \)

Now, substitute these sums back into the original expression:
\( (1^3 + 2^3 + 3^3 + \ldots + 15^3) - (1 + 2 + 3 + \ldots + 15) \)
\( = 14400 - 120 \)
\( = 14280 \)
In simple words: This problem asks you to subtract the sum of numbers from 1 to 15 from the sum of their cubes. You need to use the specific math formulas for these two sums. Calculate each sum separately and then subtract the results to find the final answer.

🎯 Exam Tip: Memorize the formulas for the sum of the first 'n' natural numbers \( (\Sigma n) \) and the sum of their cubes \( (\Sigma n^3) \). These are very common in series problems and save a lot of calculation time.