Samacheer Kalvi Class 10 Maths Solutions Chapter 2 Numbers and Sequences Exercise 2.9

Question 1. Find the sum of the following series
(i) \( 1 + 2 + 3 + ....... + 60 \)
(ii) \( 3 + 6 + 9 + ....... + 96 \)
(iii) \( 51 + 52 + 53 + ....... + 92 \)
(iv) \( 1 + 4 + 9 + 16 + ....... + 225 \)
(v) \( 6^2 + 7^2 + 8^2 + ....... + 21^2 \)
(vi) \( 10^3 + 11^3 + 12^3 + ....... + 20^3 \)
(vii) \( 1 + 3 + 5 + ...... + 71 \)
Answer:
(i) We need to find the sum of the first 60 natural numbers. The formula for the sum of the first \( n \) natural numbers is \( \frac{n(n+1)}{2} \).
So, for \( n = 60 \):
Sum \( = \frac{60 \times (60+1)}{2} \)
\( = \frac{60 \times 61}{2} \)
\( = 30 \times 61 \)
\( = 1830 \)

(ii) This series is \( 3 + 6 + 9 + ....... + 96 \). We can take 3 common from each term.
\( = 3(1 + 2 + 3 + ....... + 32) \)
Now, we find the sum of the first 32 natural numbers using the formula \( \frac{n(n+1)}{2} \).
Sum \( = 3 \times \frac{32 \times (32+1)}{2} \)
\( = 3 \times \frac{32 \times 33}{2} \)
\( = 3 \times 16 \times 33 \)
\( = 48 \times 33 \)
\( = 1584 \)

(iii) This series is \( 51 + 52 + 53 + ....... + 92 \). This is a sum of natural numbers starting from 51.
We can find this by taking the sum of all natural numbers up to 92 and subtracting the sum of natural numbers up to 50.
Sum \( = (1 + 2 + 3 + ....... + 92) - (1 + 2 + 3 + ....... + 50) \)
Using the formula \( \frac{n(n+1)}{2} \):
Sum up to 92 \( = \frac{92 \times (92+1)}{2} = \frac{92 \times 93}{2} = 46 \times 93 = 4278 \)
Sum up to 50 \( = \frac{50 \times (50+1)}{2} = \frac{50 \times 51}{2} = 25 \times 51 = 1275 \)
Total Sum \( = 4278 - 1275 \)
\( = 3003 \)

(iv) This series is \( 1 + 4 + 9 + 16 + ....... + 225 \). These are perfect squares: \( 1^2 + 2^2 + 3^2 + 4^2 + ....... + 15^2 \).
The last term, 225, is \( 15^2 \). So we need to sum the squares of the first 15 natural numbers.
The formula for the sum of the squares of the first \( n \) natural numbers is \( \frac{n(n+1)(2n+1)}{6} \).
For \( n = 15 \):
Sum \( = \frac{15 \times (15+1) \times (2 \times 15+1)}{6} \)
\( = \frac{15 \times 16 \times 31}{6} \)
\( = 5 \times 8 \times 31 \)
\( = 40 \times 31 \)
\( = 1240 \)

(v) This series is \( 6^2 + 7^2 + 8^2 + ....... + 21^2 \). This is a sum of squares starting from \( 6^2 \).
We can find this by taking the sum of squares up to \( 21^2 \) and subtracting the sum of squares up to \( 5^2 \).
Sum \( = (1^2 + 2^2 + 3^2 + ....... + 21^2) - (1^2 + 2^2 + 3^2 + 4^2 + 5^2) \)
Using the formula \( \frac{n(n+1)(2n+1)}{6} \):
Sum up to \( 21^2 \) (for \( n = 21 \)):
\( = \frac{21 \times (21+1) \times (2 \times 21+1)}{6} = \frac{21 \times 22 \times 43}{6} = 7 \times 11 \times 43 = 3311 \)
Sum up to \( 5^2 \) (for \( n = 5 \)):
\( = \frac{5 \times (5+1) \times (2 \times 5+1)}{6} = \frac{5 \times 6 \times 11}{6} = 5 \times 11 = 55 \)
Total Sum \( = 3311 - 55 \)
\( = 3256 \)

(vi) This series is \( 10^3 + 11^3 + 12^3 + ....... + 20^3 \). This is a sum of cubes starting from \( 10^3 \).
We can find this by taking the sum of cubes up to \( 20^3 \) and subtracting the sum of cubes up to \( 9^3 \).
Sum \( = (1^3 + 2^3 + 3^3 + ....... + 20^3) - (1^3 + 2^3 + 3^3 + ....... + 9^3) \)
The formula for the sum of the cubes of the first \( n \) natural numbers is \( \left(\frac{n(n+1)}{2}\right)^2 \).
Sum up to \( 20^3 \) (for \( n = 20 \)):
\( = \left(\frac{20 \times (20+1)}{2}\right)^2 = \left(\frac{20 \times 21}{2}\right)^2 = (10 \times 21)^2 = (210)^2 = 44100 \)
Sum up to \( 9^3 \) (for \( n = 9 \)):
\( = \left(\frac{9 \times (9+1)}{2}\right)^2 = \left(\frac{9 \times 10}{2}\right)^2 = (9 \times 5)^2 = (45)^2 = 2025 \)
Total Sum \( = 44100 - 2025 \)
\( = 42075 \)

(vii) This series is \( 1 + 3 + 5 + ...... + 71 \). This is an arithmetic progression of odd numbers.
First term \( a = 1 \), common difference \( d = 3 - 1 = 2 \), last term \( l = 71 \).
First, find the number of terms \( n \). The formula for the nth term is \( l = a + (n-1)d \).
\( 71 = 1 + (n-1)2 \)
\( 70 = (n-1)2 \)
\( 35 = n-1 \)
\( n = 36 \)
The sum of an arithmetic progression can also be found using the formula \( S_n = \frac{n}{2}(a+l) \).
\( S_{36} = \frac{36}{2}(1+71) \)
\( = 18 \times 72 \)
\( = 1296 \)
Alternatively, the sum of the first \( n \) odd numbers is \( n^2 \). Since there are 36 terms, the sum is \( 36^2 = 1296 \).
In simple words: For each series, we use a specific formula to add up the numbers. For regular numbers, squares, or cubes, there are direct formulas based on how many terms there are. For numbers that start later or skip numbers, we often find the total sum and then subtract the part we don't need. Odd number series have a special shortcut.

🎯 Exam Tip: Always identify the type of series (arithmetic, sum of squares, sum of cubes) first to choose the correct sum formula. Remember that a series starting from a number other than 1 requires subtracting a partial sum.

Question 2. If \( 1 + 2 + 3 + .... + k = 325 \), then find \( 1^3 + 2^3 + 3^3 + ....... + k^3 \)
Answer:
We are given that the sum of the first \( k \) natural numbers is 325.
So, \( 1 + 2 + 3 + .... + k = 325 \).
The formula for the sum of the first \( k \) natural numbers is \( \frac{k(k+1)}{2} \).
Therefore, we have \( \frac{k(k+1)}{2} = 325 \).
We need to find the sum of the cubes of the first \( k \) natural numbers, which is \( 1^3 + 2^3 + 3^3 + ....... + k^3 \).
The formula for the sum of the cubes of the first \( k \) natural numbers is \( \left(\frac{k(k+1)}{2}\right)^2 \).
Since we know \( \frac{k(k+1)}{2} = 325 \), we can substitute this value directly into the cubes formula.
So, \( 1^3 + 2^3 + 3^3 + ....... + k^3 = (325)^2 \)
\( = 105625 \)
In simple words: The sum of the first 'k' numbers is 325. To find the sum of the cubes of these same 'k' numbers, you simply take that sum (325) and multiply it by itself. This is because the formula for the sum of cubes is the square of the formula for the sum of natural numbers.

🎯 Exam Tip: Recognize the direct relationship between the sum of natural numbers and the sum of their cubes. If you know one, you can easily find the other without solving for 'k' itself.

Question 3. If \( 1^3 + 2^3 + 3^3 + ....... + K^3 = 44100 \) then find \( 1 + 2 + 3 + ....... + k \)
Answer:
We are given that the sum of the cubes of the first \( K \) natural numbers is 44100.
So, \( 1^3 + 2^3 + 3^3 + ....... + K^3 = 44100 \).
The formula for the sum of the cubes of the first \( K \) natural numbers is \( \left(\frac{K(K+1)}{2}\right)^2 \).
Therefore, we have \( \left(\frac{K(K+1)}{2}\right)^2 = 44100 \).
To find \( \frac{K(K+1)}{2} \), we take the square root of both sides.
\( \frac{K(K+1)}{2} = \sqrt{44100} \)
\( \frac{K(K+1)}{2} = 210 \)
We need to find the sum of the first \( K \) natural numbers, which is \( 1 + 2 + 3 + ....... + K \).
The formula for this sum is \( \frac{K(K+1)}{2} \).
So, \( 1 + 2 + 3 + ....... + K = 210 \).
In simple words: The sum of the cubes of the first 'K' numbers is 44100. To find the sum of the first 'K' numbers themselves, you just take the square root of 44100. This is the reverse of the previous problem, using the same relationship between sum of numbers and sum of cubes.

🎯 Exam Tip: Understand that \( \sqrt{X^2} = X \). When finding the sum of natural numbers from the sum of cubes, simply take the square root of the given cube sum.

Question 4. How many terms of the series \( 1^3 + 2^3 + 3^3 + ........... \) should be taken to get the sum 14400?
Answer:
We are looking for the number of terms, let's call it \( n \), such that the sum of the cubes of the first \( n \) natural numbers is 14400.
The formula for the sum of the cubes of the first \( n \) natural numbers is \( \left(\frac{n(n+1)}{2}\right)^2 \).
So, we have the equation: \( \left(\frac{n(n+1)}{2}\right)^2 = 14400 \).
First, take the square root of both sides:
\( \frac{n(n+1)}{2} = \sqrt{14400} \)
\( \frac{n(n+1)}{2} = 120 \)
Now, multiply both sides by 2:
\( n(n+1) = 120 \times 2 \)
\( n(n+1) = 240 \)
Expand the left side:
\( n^2 + n = 240 \)
Rearrange it into a quadratic equation:
\( n^2 + n - 240 = 0 \)
We need to find two numbers that multiply to -240 and add to 1. These numbers are 16 and -15.
So, we can factor the quadratic equation:
\( (n + 16)(n - 15) = 0 \)
This gives two possible values for \( n \):
\( n + 16 = 0 \implies n = -16 \)
\( n - 15 = 0 \implies n = 15 \)
Since the number of terms cannot be negative, we disregard \( n = -16 \).
Therefore, the number of terms taken is \( 15 \).
In simple words: We are looking for how many numbers, when cubed and added together, make 14400. We use the formula for summing cubes and work backwards. First, we find the square root of 14400, then we solve a simple equation to find the count of numbers. Since you can't have a negative number of terms, we pick the positive answer.

🎯 Exam Tip: When solving for 'n' in sequence problems, always consider that 'n' must be a positive integer, as it represents a count of terms. Discard any negative or non-integer solutions.

Question 5. The sum of the squares of the first n natural numbers is 285, while the sum of their cubes is 2025. Find the value of n.
Answer:
We are given two pieces of information:
1. Sum of squares of the first \( n \) natural numbers \( = 285 \)
The formula for sum of squares is \( \frac{n(n+1)(2n+1)}{6} = 285 \). (Equation 1)
2. Sum of cubes of the first \( n \) natural numbers \( = 2025 \)
The formula for sum of cubes is \( \left(\frac{n(n+1)}{2}\right)^2 = 2025 \). (Equation 2)

Let's use Equation 2 first, as it's simpler to solve for \( \frac{n(n+1)}{2} \).
\( \left(\frac{n(n+1)}{2}\right)^2 = 2025 \)
Take the square root of both sides:
\( \frac{n(n+1)}{2} = \sqrt{2025} \)
\( \frac{n(n+1)}{2} = 45 \). (Equation 3)

Now, substitute the value \( \frac{n(n+1)}{2} = 45 \) into Equation 1.
Equation 1 can be rewritten as \( \frac{n(n+1)}{2} \times \frac{2n+1}{3} = 285 \).
Substitute 45 for \( \frac{n(n+1)}{2} \):
\( 45 \times \frac{2n+1}{3} = 285 \)
\( 15 \times (2n+1) = 285 \)
Divide both sides by 15:
\( 2n+1 = \frac{285}{15} \)
\( 2n+1 = 19 \)
Subtract 1 from both sides:
\( 2n = 19 - 1 \)
\( 2n = 18 \)
Divide by 2:
\( n = \frac{18}{2} \)
\( n = 9 \)
So, the value of \( n \) is 9.
In simple words: We are given the sum of squares and the sum of cubes for the same number of terms, 'n'. We use the sum of cubes formula first, as it quickly gives us the value of \( \frac{n(n+1)}{2} \). Then we plug this value into the sum of squares formula. This helps us solve for 'n' directly.

🎯 Exam Tip: When given multiple related sums (like sum of squares and sum of cubes), start with the formula that allows for easier isolation of a common term, like \( \frac{n(n+1)}{2} \), which is present in both sum of numbers and sum of cubes formulas.

Question 6. Rekha has 15 square colour papers of sizes 10 cm, 11 cm, 12 cm, ..., 24 cm. How much area can be decorated with these colour papers?
Answer:
To find the total area that can be decorated, we need to find the sum of the areas of all the square papers. The area of a square is \( \text{side}^2 \).
The sides of the square papers are 10 cm, 11 cm, 12 cm, ..., 24 cm.
So, we need to calculate the sum of squares: \( 10^2 + 11^2 + 12^2 + ....... + 24^2 \).
This is a sum of squares starting from \( 10^2 \). We can find this by taking the sum of squares up to \( 24^2 \) and subtracting the sum of squares up to \( 9^2 \).
Sum \( = (1^2 + 2^2 + 3^2 + ....... + 24^2) - (1^2 + 2^2 + 3^2 + ....... + 9^2) \)
The formula for the sum of the squares of the first \( n \) natural numbers is \( \frac{n(n+1)(2n+1)}{6} \).
Sum up to \( 24^2 \) (for \( n = 24 \)):
\( = \frac{24 \times (24+1) \times (2 \times 24+1)}{6} = \frac{24 \times 25 \times 49}{6} \)
\( = 4 \times 25 \times 49 = 100 \times 49 = 4900 \)
Sum up to \( 9^2 \) (for \( n = 9 \)):
\( = \frac{9 \times (9+1) \times (2 \times 9+1)}{6} = \frac{9 \times 10 \times 19}{6} \)
\( = 3 \times 5 \times 19 = 15 \times 19 = 285 \)
Total Area \( = 4900 - 285 \)
\( = 4615 \)
Therefore, the area that can be decorated is 4615 cm\(^2\).
In simple words: To find the total decorated area, we need to add up the area of each square paper. Since the papers have different side lengths, we calculate the square of each side and add them all up. We use a formula to quickly sum squares, but since the sizes start from 10 cm, we calculate the sum up to 24 cm and remove the sum for sizes 1 cm to 9 cm.

🎯 Exam Tip: Remember to calculate the total sum of squares and then subtract the sum of squares of the missing initial terms. Also, ensure units are correctly mentioned in the final answer.

Question 7. Find the sum of the series \( (2^3 - 1^3)+(4^3 - 3^3) + (6^3 - 5^3) + ........ \) to
(i) n terms
(ii) 8 terms
Answer:
(i) For \( n \) terms, the series is \( (2^3 - 1^3)+(4^3 - 3^3) + (6^3 - 5^3) + \dots + ((2n)^3 - (2n-1)^3) \).
This can be written as the sum of even cubes minus the sum of odd cubes up to \( (2n-1)^3 \).
Sum of the series \( = (2^3 + 4^3 + 6^3 + \dots + (2n)^3) - (1^3 + 3^3 + 5^3 + \dots + (2n-1)^3) \)
Let's find the sum of \( (2^3 + 4^3 + 6^3 + \dots + (2n)^3) \):
\( = 2^3(1^3 + 2^3 + 3^3 + \dots + n^3) \)
\( = 8 \times \left(\frac{n(n+1)}{2}\right)^2 \)
\( = 8 \times \frac{n^2(n+1)^2}{4} = 2n^2(n+1)^2 \). (Equation 1)

Now, let's find the sum of \( (1^3 + 3^3 + 5^3 + \dots + (2n-1)^3) \).
We know that Sum of first \( (2n) \) natural cubes \( = (1^3 + 2^3 + 3^3 + \dots + (2n)^3) \).
And Sum of first \( n \) even cubes \( = 2n^2(n+1)^2 \).
So, Sum of first \( n \) odd cubes \( = \) (Sum of first \( 2n \) natural cubes) - (Sum of first \( n \) even cubes)
\( = \left(\frac{2n(2n+1)}{2}\right)^2 - 2n^2(n+1)^2 \)
\( = (n(2n+1))^2 - 2n^2(n+1)^2 \)
\( = n^2(2n+1)^2 - 2n^2(n+1)^2 \)
\( = n^2[(2n+1)^2 - 2(n+1)^2] \)
\( = n^2[(4n^2 + 4n + 1) - 2(n^2 + 2n + 1)] \)
\( = n^2[4n^2 + 4n + 1 - 2n^2 - 4n - 2] \)
\( = n^2[2n^2 - 1] \)
\( = 2n^4 - n^2 \). (Equation 2)

Now, substitute Equation 1 and Equation 2 back into the original sum of the series:
Sum \( = 2n^2(n+1)^2 - (2n^4 - n^2) \)
\( = 2n^2(n^2 + 2n + 1) - 2n^4 + n^2 \)
\( = 2n^4 + 4n^3 + 2n^2 - 2n^4 + n^2 \)
\( = 4n^3 + 3n^2 \)

(ii) For 8 terms, we substitute \( n = 8 \) into the formula derived in part (i):
Sum \( = 4(8)^3 + 3(8)^2 \)
\( = 4(512) + 3(64) \)
\( = 2048 + 192 \)
\( = 2240 \)
In simple words: For part (i), we split the series into two parts: the sum of cubes of even numbers and the sum of cubes of odd numbers. We found a formula for each and then subtracted the odd sum from the even sum to get a general formula in terms of 'n'. For part (ii), we just used this new general formula and put 8 in place of 'n' to find the total sum for 8 terms.

🎯 Exam Tip: When dealing with series involving sums of even or odd numbers/powers, often it's helpful to express them in terms of sums of natural numbers by factoring out common terms (like \( 2^3 \) for even series) or by using subtraction (total sum minus even sum gives odd sum).