Samacheer Kalvi Class 10 Maths Solutions Chapter 2 Numbers and Sequences Exercise 2

Question 1. Prove that \( n^2 - n \) is divisible by 2 for every positive integer n.
Answer: We know that any positive integer n can be written in two forms: either \( 2q \) (an even number) or \( 2q + 1 \) (an odd number) for some integer q. Let's check both cases to see if \( n^2 - n \) is always divisible by 2.

Case 1: When \( n = 2q \)

\( n^2 - n = (2q)^2 - 2q \)
\( = 4q^2 - 2q \)
\( = 2q(2q - 1) \)

Here, since the expression is \( 2q(2q - 1) \), it is a multiple of 2. So, \( n^2 - n \) is divisible by 2. We can call \( q(2q-1) \) as 'r', so the expression is \( 2r \).

Case 2: When \( n = 2q + 1 \)

\( n^2 - n = (2q + 1)^2 - (2q + 1) \)
\( = (4q^2 + 4q + 1) - (2q + 1) \)
\( = 4q^2 + 4q + 1 - 2q - 1 \)
\( = 4q^2 + 2q \)
\( = 2q(2q + 1) \)

Again, since the expression is \( 2q(2q + 1) \), it is a multiple of 2. So, \( n^2 - n \) is divisible by 2. We can call \( q(2q+1) \) as 'r', so the expression is \( 2r \). In both cases, \( n^2 - n \) can be written as \( 2r \), which means it is always an even number and thus divisible by 2.
In simple words: No matter if 'n' is an even or odd number, when you calculate \( n^2 - n \), the result will always be an even number. This means it can always be divided by 2 without any remainder.

🎯 Exam Tip: When proving divisibility for all positive integers, always consider both even and odd cases by representing n as 2q and 2q+1, respectively.

Question 2. A milkman has 175 litres of cow's milk and 105 litres of buffalo's milk. He wishes to sell the milk by filling the two types of milk in cans of equal capacity. Calculate the following:
(i) Capacity of a can
(ii) Number of cans of cow's milk
(iii) Number of cans of buffalo's milk.

Answer: To find the largest equal capacity for the cans, we need to find the Highest Common Factor (H.C.F.) of 175 litres and 105 litres. We will use Euclid's division algorithm for this.

First, we divide 175 by 105:
\( 175 = 105 \times 1 + 70 \) (The remainder is 70, which is not 0.)

Next, we divide 105 by the remainder 70:
\( 105 = 70 \times 1 + 35 \) (The remainder is 35, which is not 0.)

Finally, we divide 70 by the remainder 35:
\( 70 = 35 \times 2 + 0 \) (The remainder is 0.)

Since the remainder is now 0, the H.C.F. is the last non-zero divisor, which is 35.

(i) The milkman's milk can's capacity should be 35 litres to hold both types of milk in equal measures.

(ii) Number of cans for cow's milk \( = \frac{175}{35} = 5 \) cans.

(iii) Number of cans for buffalo's milk \( = \frac{105}{35} = 3 \) cans.
In simple words: To find the largest cans that can hold both types of milk equally, we find the biggest number that divides both 175 and 105. That number is 35. So, each can will hold 35 litres. You will need 5 cans for cow's milk and 3 cans for buffalo's milk.

🎯 Exam Tip: Remember that "equal capacity" questions often involve finding the H.C.F. (Highest Common Factor) using Euclid's division algorithm to determine the largest possible size.

Question 3. When the positive integers a, b and c are divided by 13, the respective remainders are 9, 7 and 10. Find the remainder when a + 2b + 3c is divided by 13.
Answer: We are given that when positive integers a, b, and c are divided by 13, they leave remainders 9, 7, and 10 respectively. We can write this using the division algorithm:

\( a = 13q_1 + 9 \)
\( b = 13q_2 + 7 \)
\( c = 13q_3 + 10 \)

Now, let's find \( a + 2b + 3c \):
\( a + 2b + 3c = (13q_1 + 9) + 2(13q_2 + 7) + 3(13q_3 + 10) \)
\( = 13q_1 + 9 + 26q_2 + 14 + 39q_3 + 30 \)
\( = 13q_1 + 26q_2 + 39q_3 + 9 + 14 + 30 \)
\( = 13(q_1 + 2q_2 + 3q_3) + 53 \)

Let \( Q = q_1 + 2q_2 + 3q_3 \). Then, \( a + 2b + 3c = 13Q + 53 \). To find the remainder when this sum is divided by 13, we need to find the remainder of 53 when divided by 13. The expression is in the form \( 13 \times \text{some_integer} + 53 \).
The remainder as per the solution given is 53. However, in standard mathematics, the remainder should be less than the divisor (13). If we perform the division \( 53 \div 13 \), we get \( 53 = 13 \times 4 + 1 \). So, the standard remainder would be 1. The original solution provided a remainder of 53.
In simple words: When we add up 'a', 'b', and 'c' (with 'b' multiplied by 2 and 'c' by 3), the total number can be written as a multiple of 13 plus 53. If you were to divide this total by 13, the solution indicates the remainder is 53.

🎯 Exam Tip: When working with remainders, remember that if the remainder is larger than the divisor, you should divide it again by the divisor to find the final, smallest non-negative remainder.

Question 4. Show that 107 is of the form 4q +3 for any integer q.
Answer: To show that 107 can be written in the form \( 4q + 3 \), we need to divide 107 by 4 and find the quotient (q) and the remainder (r).
When we divide 107 by 4:
\( 107 = 4 \times 26 + 3 \)

Here, the quotient \( q = 26 \) and the remainder \( r = 3 \). This matches the form \( a = bq + r \), where \( a = 107 \), \( b = 4 \), \( q = 26 \), and \( r = 3 \). This proves that 107 is indeed of the form \( 4q + 3 \).
In simple words: If you divide 107 by 4, you get 26 with 3 left over. This shows that 107 can be written as "4 times some number, plus 3".

🎯 Exam Tip: To show a number is of the form \( bq + r \), simply perform the division of the number by 'b' and identify the quotient as 'q' and the remainder as 'r'.

Question 5. If \( (m + 1)^{th} \) term of an A.P. is twice the \( (n + 1)^{th} \) term, then prove that \( (3m + 1)^{th} \) term is twice the \( (m + n + 1)^{th} \) term.
Answer: Let 'a' be the first term and 'd' be the common difference of the Arithmetic Progression (A.P.). The formula for the \( n^{th} \) term of an A.P. is \( t_n = a + (n-1)d \).

We are given that the \( (m+1)^{th} \) term is twice the \( (n+1)^{th} \) term:
\( t_{m+1} = 2t_{n+1} \)
\( a + ((m+1) - 1)d = 2[a + ((n+1) - 1)d] \)
\( a + md = 2[a + nd] \)
\( \implies a + md = 2a + 2nd \)
Rearranging the terms to find a relationship for 'a':
\( md - 2nd = 2a - a \)
\( d(m - 2n) = a \) ....(1)

Now we need to prove that the \( (3m+1)^{th} \) term is twice the \( (m+n+1)^{th} \) term. First, let's find the Left Hand Side (L.H.S.), which is \( t_{3m+1} \):
\( L.H.S. = t_{3m+1} \)
\( = a + ((3m+1) - 1)d \)
\( = a + 3md \)

Substitute the value of 'a' from equation (1) into this expression:
\( = d(m - 2n) + 3md \)
\( = md - 2nd + 3md \)
\( = 4md - 2nd \)
\( = 2d(2m - n) \)

Next, let's find the Right Hand Side (R.H.S.), which is \( 2t_{m+n+1} \):
\( R.H.S. = 2t_{m+n+1} \)
\( = 2[a + ((m+n+1) - 1)d] \)
\( = 2[a + (m+n)d] \)

Substitute the value of 'a' from equation (1) into this expression:
\( = 2[d(m - 2n) + (m+n)d] \)
\( = 2[md - 2nd + md + nd] \)
\( = 2[2md - nd] \)
\( = 2d(2m - n) \)

Since L.H.S. \( = 2d(2m - n) \) and R.H.S. \( = 2d(2m - n) \), we have L.H.S. = R.H.S. Therefore, \( t_{3m+1} = 2t_{m+n+1} \). This proves the statement.
In simple words: We used the formula for terms in a number pattern (Arithmetic Progression). First, we used the given information to find a link between the first term 'a' and the common difference 'd'. Then, we used this link to show that the \( (3m+1)^{th} \) term is indeed double the \( (m+n+1)^{th} \) term, just as the problem asked.

🎯 Exam Tip: For A.P. proofs, always start by defining 'a' (first term) and 'd' (common difference). Clearly write out the expressions for each term involved and use substitution to simplify both sides of the equation you need to prove.

Question 6. Find the \( 12^{th} \) term from the last term of the A.P. -2, -4, -6,... -100.
Answer: The given Arithmetic Progression (A.P.) is -2, -4, -6,..., -100. To find a term from the last, it's easier to reverse the A.P. and find the \( 12^{th} \) term from the beginning of the reversed A.P.

In the original A.P.:
First term \( a = -2 \)
Common difference \( d = -4 - (-2) = -4 + 2 = -2 \)

When we reverse the A.P., the new first term (which is the original last term) becomes:
New \( a = -100 \)
The common difference changes its sign: New \( d = -(-2) = 2 \)

Now we need to find the \( 12^{th} \) term of this new A.P. using the formula \( t_n = a + (n-1)d \):
\( t_{12} = -100 + (12-1)(2) \)
\( = -100 + 11(2) \)
\( = -100 + 22 \)
\( = -78 \)

So, the \( 12^{th} \) term from the last term of the given A.P. is -78.
In simple words: To find a term from the end of a list of numbers, you can just flip the list around. The last number becomes the first, and the common difference changes sign. Then, you find the requested term (here, the 12th) from the start of this new list.

🎯 Exam Tip: When calculating terms from the end of an A.P., remember to reverse the series and change the sign of the common difference before applying the standard \( t_n \) formula.

Question 7. Two A.P.'s have the same common difference. The first term of one A.P. is 2 and that of the other is 7. Show that the difference between their \( 10^{th} \) terms is the same as the difference between their \( 21^{st} \) terms, which is the same as the difference between any two corresponding terms.
Answer: Let the common difference for both A.P.'s be 'd'.

For the first A.P. (\( AP_1 \)):
First term \( a_1 = 2 \)
The terms are: \( a_1, a_1 + d, a_1 + 2d, \dots \)

For the second A.P. (\( AP_2 \)):
First term \( a_2 = 7 \)
The terms are: \( a_2, a_2 + d, a_2 + 2d, \dots \)

Now, let's find the difference between their \( 10^{th} \) terms:
\( t_{10} \) in \( AP_1 = a_1 + (10-1)d = 2 + 9d \)
\( t_{10} \) in \( AP_2 = a_2 + (10-1)d = 7 + 9d \)

Difference between their \( 10^{th} \) terms:
\( (2 + 9d) - (7 + 9d) = 2 + 9d - 7 - 9d = -5 \)

Next, let's find the difference between their \( 21^{st} \) terms:
\( t_{21} \) in \( AP_1 = a_1 + (21-1)d = 2 + 20d \)
\( t_{21} \) in \( AP_2 = a_2 + (21-1)d = 7 + 20d \)

Difference between their \( 21^{st} \) terms:
\( (2 + 20d) - (7 + 20d) = 2 + 20d - 7 - 20d = -5 \)

As we can see, the difference between their \( 10^{th} \) terms is -5, and the difference between their \( 21^{st} \) terms is also -5. These differences are the same.

Now, let's consider the difference between any two corresponding \( k^{th} \) terms:
\( t_k \) in \( AP_1 = a_1 + (k-1)d = 2 + (k-1)d \)
\( t_k \) in \( AP_2 = a_2 + (k-1)d = 7 + (k-1)d \)

Difference between their \( k^{th} \) terms:
\( (2 + (k-1)d) - (7 + (k-1)d) = 2 + (k-1)d - 7 - (k-1)d = -5 \)

This shows that the difference between any two corresponding terms is always -5, which is constant. This happens because the common difference 'd' cancels out in the subtraction. Therefore, the statement is proven.
In simple words: When two sets of numbers (A.P.s) grow at the same speed (same common difference), the gap between their matching numbers (like the 10th term of one and the 10th term of the other) will always stay the same. In this case, because the second list started 5 higher, its numbers will always be 5 higher than the first list's numbers.

🎯 Exam Tip: When the common difference is the same for two A.P.s, the difference between their corresponding terms is always constant and equal to the difference between their first terms.

Question 8. A man saved Rs 16500 in ten years. In each year after the first, he saved Rs 100 more than he did in the preceding year. How much did he save in the first year?
Answer: This problem involves an Arithmetic Progression (A.P.) because the man's savings increase by a fixed amount each year.

Let 'a' be the amount saved in the first year.
The increase in saving each year is Rs 100, so the common difference \( d = 100 \).
The total amount saved in ten years is Rs 16500, so \( S_{10} = 16500 \).
The number of years \( n = 10 \).

We use the formula for the sum of the first 'n' terms of an A.P.:
\( S_n = \frac{n}{2} [2a + (n - 1)d] \)

Substitute the given values:
\( S_{10} = \frac{10}{2} [2a + (10 - 1)100] \)
\( 16500 = 5 [2a + 9 \times 100] \)
\( 16500 = 5 [2a + 900] \)

Divide both sides by 5:
\( \frac{16500}{5} = 2a + 900 \)
\( 3300 = 2a + 900 \)

Subtract 900 from both sides:
\( 3300 - 900 = 2a \)
\( 2400 = 2a \)

Divide by 2 to find 'a':
\( a = \frac{2400}{2} \)
\( a = 1200 \)

So, the amount saved in the first year was Rs 1200.
In simple words: The man saved money in a pattern where he saved an extra Rs 100 each year. Over 10 years, he saved a total of Rs 16500. Using a special math rule for such patterns, we found that he must have saved Rs 1200 in the very first year.

🎯 Exam Tip: Always identify the first term (a), common difference (d), and number of terms (n) first in A.P. word problems. Then, choose the correct formula for sum (S_n) or term (t_n).

Question 9. Find the G.P. in which the \( 2^{nd} \) term is \( \sqrt{6} \) and the \( 6^{th} \) term is \( 9\sqrt{6} \).
Answer: Let 'a' be the first term and 'r' be the common ratio of the Geometric Progression (G.P.). The formula for the \( n^{th} \) term of a G.P. is \( t_n = ar^{n-1} \).

We are given the \( 2^{nd} \) term:
\( t_2 = ar^{2-1} = ar = \sqrt{6} \) ....(1)

We are also given the \( 6^{th} \) term:
\( t_6 = ar^{6-1} = ar^5 = 9\sqrt{6} \) ....(2)

To find 'r', divide equation (2) by equation (1):
\( \frac{ar^5}{ar} = \frac{9\sqrt{6}}{\sqrt{6}} \)
\( r^4 = 9 \)

To find 'r', take the fourth root of 9:
\( r^4 = (\sqrt{3})^4 \)
\( \implies r = \sqrt{3} \)

Now, substitute the value of \( r = \sqrt{3} \) back into equation (1) to find 'a':
\( ar = \sqrt{6} \)
\( a \times \sqrt{3} = \sqrt{6} \)
\( a = \frac{\sqrt{6}}{\sqrt{3}} \)
\( a = \sqrt{\frac{6}{3}} \)
\( a = \sqrt{2} \)

So, the first term is \( a = \sqrt{2} \) and the common ratio is \( r = \sqrt{3} \). The G.P. is \( a, ar, ar^2, \dots \)
Substituting the values:
\( \sqrt{2}, \quad \sqrt{2} \times \sqrt{3}, \quad \sqrt{2} \times (\sqrt{3})^2, \dots \)
\( \sqrt{2}, \quad \sqrt{6}, \quad \sqrt{2} \times 3, \dots \)
\( \sqrt{2}, \quad \sqrt{6}, \quad 3\sqrt{2}, \dots \)

The Geometric Progression is \( \sqrt{2}, \sqrt{6}, 3\sqrt{2}, \dots \).
In simple words: We used the given terms to find the starting number (first term) and the multiplication factor (common ratio) of the geometric pattern. By dividing the 6th term by the 2nd term, we found the common ratio. Then, using the 2nd term, we found the first term. Finally, we listed the first few numbers in the pattern.

🎯 Exam Tip: For G.P. problems with two given terms, dividing one term by another (especially \( t_n / t_m \)) is an effective way to find the common ratio 'r' first. Always clearly state 'a' and 'r' before writing out the G.P.

Question 10. The value of a motorcycle depreciates at a rate of 15% per year. What will be the value of the motorcycle 3 years hence, which is now purchased for Rs 45,000?
Answer: This problem involves depreciation, meaning the value decreases each year. We need to calculate the value after 3 years.

Initial value of the motorcycle (a) = Rs 45,000.
Depreciation rate = 15% per year.

**Year 1:**
Depreciation in the 1st year = 15% of Rs 45,000
\( = 45000 \times \frac{15}{100} = 450 \times 15 = Rs 6750 \)

Value at the end of the 1st year = Initial value - Depreciation
\( = Rs 45000 - Rs 6750 = Rs 38250 \)

**Year 2:**
Depreciation in the 2nd year = 15% of the value at the end of the 1st year
\( = 38250 \times \frac{15}{100} = 382.50 \times 15 = Rs 5737.50 \)

Value at the end of the 2nd year = Value after 1st year - Depreciation
\( = Rs 38250 - Rs 5737.50 = Rs 32512.50 \)

**Year 3:**
Depreciation in the 3rd year = 15% of the value at the end of the 2nd year
\( = 32512.50 \times \frac{15}{100} = 325.125 \times 15 = Rs 4876.875 \)

Value at the end of the 3rd year = Value after 2nd year - Depreciation
\( = Rs 32512.50 - Rs 4876.875 = Rs 27635.625 \)

Rounding to the nearest rupee, the value of the motorcycle after 3 years will be approximately Rs 27636.
In simple words: The motorcycle loses 15% of its value each year. We calculated this loss for three years, one year at a time. After three years, starting from Rs 45,000, its value will be around Rs 27636. This type of reduction is like compound interest, but in reverse.

🎯 Exam Tip: For depreciation problems, always calculate the percentage loss on the *current* value, not the original value, for each successive period. Be careful with calculations involving decimals and remember to round the final answer appropriately if specified.