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

10th Maths Chapter 2: Numbers and Sequences

Exercise 2.9 Solutions

Important Formulae

• Sum of first n natural numbers: $$ 1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2} $$
• Sum of first n odd natural numbers: $$ 1 + 3 + 5 + \dots + (2n-1) = n^2 $$
• Sum of squares of first n natural numbers: $$ 1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6} $$
• Sum of cubes of first n natural numbers: $$ 1^3 + 2^3 + 3^3 + \dots + n^3 = \left[ \frac{n(n+1)}{2} \right]^2 $$

Question 1

Find the sum of the following series

(i) \( 1 + 2 + 3 + \dots + 60 \)

This is the sum of the first 60 natural numbers. Here, \( n = 60 \).

Using the formula: $$ \text{Sum} = \frac{n(n+1)}{2} $$

$$ \text{Sum} = \frac{60(60+1)}{2} = \frac{60(61)}{2} = 30 \times 61 $$

Sum = 1830

(ii) \( 3 + 6 + 9 + \dots + 96 \)

First, take the common factor 3 out of the series.

$$ \text{Sum} = 3(1 + 2 + 3 + \dots + 32) $$

Now, we find the sum of the first 32 natural numbers (\( n=32 \)).

$$ \text{Sum} = 3 \left[ \frac{32(32+1)}{2} \right] = 3 \left[ \frac{32(33)}{2} \right] = 3(16 \times 33) = 3(528) $$

Sum = 1584

(iii) \( 51 + 52 + 53 + \dots + 92 \)

To find this sum, we calculate the sum from 1 to 92 and subtract the sum from 1 to 50.

$$ \text{Sum} = (1 + 2 + \dots + 92) - (1 + 2 + \dots + 50) $$

$$ \text{Sum} = \left[ \frac{92(92+1)}{2} \right] - \left[ \frac{50(50+1)}{2} \right] $$

$$ \text{Sum} = \left[ \frac{92(93)}{2} \right] - \left[ \frac{50(51)}{2} \right] = (46 \times 93) - (25 \times 51) $$

$$ \text{Sum} = 4278 - 1275 $$

Sum = 3003

(iv) \( 1 + 4 + 9 + 16 + \dots + 225 \)

The series can be written as the sum of squares.

$$ \text{Sum} = 1^2 + 2^2 + 3^2 + 4^2 + \dots + 15^2 $$

This is the sum of the squares of the first 15 natural numbers (\( n=15 \)).

Using the formula: $$ \text{Sum} = \frac{n(n+1)(2n+1)}{6} $$

$$ \text{Sum} = \frac{15(15+1)(2 \times 15+1)}{6} = \frac{15(16)(31)}{6} = 5 \times 8 \times 31 $$

Sum = 1240

(v) \( 6^2 + 7^2 + 8^2 + \dots + 21^2 \)

$$ \text{Sum} = (1^2 + 2^2 + \dots + 21^2) - (1^2 + 2^2 + \dots + 5^2) $$

Part 1 (n=21): $$ \frac{21(21+1)(2 \times 21+1)}{6} = \frac{21(22)(43)}{6} = 7 \times 11 \times 43 = 3311 $$

Part 2 (n=5): $$ \frac{5(5+1)(2 \times 5+1)}{6} = \frac{5(6)(11)}{6} = 55 $$

Total Sum = 3311 - 55

Sum = 3256

(vi) \( 10^3 + 11^3 + 12^3 + \dots + 20^3 \)

$$ \text{Sum} = (1^3 + 2^3 + \dots + 20^3) - (1^3 + 2^3 + \dots + 9^3) $$

Part 1 (n=20): $$ \left[ \frac{20(20+1)}{2} \right]^2 = [10 \times 21]^2 = 210^2 = 44100 $$

Part 2 (n=9): $$ \left[ \frac{9(9+1)}{2} \right]^2 = [9 \times 5]^2 = 45^2 = 2025 $$

Total Sum = 44100 - 2025

Sum = 42075

(vii) \( 1 + 3 + 5 + \dots + 71 \)

This is a sum of odd numbers. The formula is \( n^2 \).

First, find the number of terms, n. Using the AP formula: $$ n = \frac{l - a}{d} + 1 $$

Here, last term \(l = 71\), first term \(a = 1\), common difference \(d = 2\).

$$ n = \frac{71 - 1}{2} + 1 = \frac{70}{2} + 1 = 35 + 1 = 36 $$

Sum = \( n^2 = 36^2 \)

Sum = 1296

Question 2

If \( 1 + 2 + 3 + \dots + k = 325 \), then find \( 1^3 + 2^3 + 3^3 + \dots + k^3 \).

We are given that the sum of the first k natural numbers is 325.

We need to find the sum of the cubes of the first k natural numbers.

The formula for the sum of cubes is: $$ 1^3 + 2^3 + \dots + k^3 = \left[ \frac{k(k+1)}{2} \right]^2 $$

We know that \( \frac{k(k+1)}{2} = 1 + 2 + \dots + k \).

Therefore, \( 1^3 + 2^3 + \dots + k^3 = (1 + 2 + \dots + k)^2 \).

Substituting the given value:

Sum of cubes = \( (325)^2 \)

Sum of cubes = 105625

Question 3

If \( 1^3 + 2^3 + 3^3 + \dots + k^3 = 44100 \), then find \( 1 + 2 + 3 + \dots + k \).

We are given the sum of the cubes of the first k natural numbers.

We know the relationship: \( 1^3 + 2^3 + \dots + k^3 = (1 + 2 + \dots + k)^2 \).

So, \( (1 + 2 + \dots + k)^2 = 44100 \).

To find the sum, we take the square root of both sides.

$$ 1 + 2 + \dots + k = \sqrt{44100} $$

Since \( 210^2 = 44100 \),

\( 1 + 2 + 3 + \dots + k = 210 \)

Question 4

How many terms of the series \( 1^3 + 2^3 + 3^3 + \dots \) should be taken to get the sum 14400?

Let the number of terms be n. The sum is given as \( S_n = 14400 \).

$$ S_n = 1^3 + 2^3 + 3^3 + \dots + n^3 = \left[ \frac{n(n+1)}{2} \right]^2 $$

$$ \left[ \frac{n(n+1)}{2} \right]^2 = 14400 $$

We know that \( 120^2 = 14400 \). So, we can equate the bases.

$$ \frac{n(n+1)}{2} = 120 $$

$$ n(n+1) = 240 \implies n^2 + n - 240 = 0 $$

Factoring the quadratic equation: $$ (n + 16)(n - 15) = 0 $$

This gives two possible values for n: \( n = -16 \) or \( n = 15 \).

Since the number of terms cannot be negative, we discard \( n = -16 \).

n = 15. So, 15 terms should be added.

Question 5

The sum of the cubes of the first n natural numbers is 2025, then find the value of n.

Given: \( 1^3 + 2^3 + 3^3 + \dots + n^3 = 2025 \).

Using the formula for the sum of cubes: $$ \left[ \frac{n(n+1)}{2} \right]^2 = 2025 $$

We know that \( 45^2 = 2025 \). Equating the bases:

$$ \frac{n(n+1)}{2} = 45 \implies n(n+1) = 90 $$

$$ n^2 + n - 90 = 0 $$

Factoring the equation: $$ (n + 10)(n - 9) = 0 $$

This gives \( n = -10 \) or \( n = 9 \).

The number of terms (n) cannot be negative.

The value of n is 9.

Question 6

Reka 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?

The area of a square is \( (\text{side})^2 \). The total area is the sum of the areas of all square papers.

$$ \text{Total Area} = 10^2 + 11^2 + 12^2 + \dots + 24^2 $$

To find this sum, we use the formula for the sum of squares:

$$ \text{Total Area} = (1^2 + 2^2 + \dots + 24^2) - (1^2 + 2^2 + \dots + 9^2) $$

Part 1 (n=24): $$ \frac{24(24+1)(2 \times 24+1)}{6} = \frac{24(25)(49)}{6} = 4 \times 25 \times 49 = 4900 $$

Part 2 (n=9): $$ \frac{9(9+1)(2 \times 9+1)}{6} = \frac{9(10)(19)}{6} = 3 \times 5 \times 19 = 285 $$

Total Area = 4900 - 285

Total Area = 4615 sq. cm.

Question 7

Find the sum of the series \( (2^3 – 1^3) + (4^3 – 3^3) + (6^3 – 5^3) + \dots \) to

(i) n terms

Let \( S_n \) be the sum of the series to n terms.

The n^th term of the series is \( (2n)^3 - (2n-1)^3 \).

Using the identity \( a^3 - b^3 = (a-b)(a^2 + ab + b^2) \). A simpler approach is to find the sum of even cubes and subtract the sum of odd cubes.

$$ S_n = \sum_{k=1}^{n} \left[ (2k)^3 - (2k-1)^3 \right] $$ The n^th term, \( t_n = (2n)^3 - (2n-1)^3 = 8n^3 - (8n^3 - 12n^2 + 6n - 1) = 12n^2 - 6n + 1 \).

The sum \( S_n \) is the summation of the n^th term:

$$ S_n = \sum_{n=1}^{k} (12n^2 - 6n + 1) = 12\sum n^2 - 6\sum n + \sum 1 $$

Using the standard summation formulas:

$$ S_n = 12 \left[ \frac{n(n+1)(2n+1)}{6} \right] - 6 \left[ \frac{n(n+1)}{2} \right] + n $$

$$ S_n = 2n(n+1)(2n+1) - 3n(n+1) + n $$

$$ S_n = 2n(2n^2 + 3n + 1) - (3n^2 + 3n) + n $$

$$ S_n = 4n^3 + 6n^2 + 2n - 3n^2 - 3n + n $$

Combining like terms:

$$ S_n = 4n^3 + (6-3)n^2 + (2-3+1)n $$

$$ S_n = 4n^3 + 3n^2 $$

(ii) 8 terms

We use the formula for the sum to n terms derived in part (i): \( S_n = 4n^3 + 3n^2 \).

Substitute \( n = 8 \) to find the sum of the first 8 terms.

$$ S_8 = 4(8)^3 + 3(8)^2 $$

$$ S_8 = 4(512) + 3(64) = 2048 + 192 $$

$$ S_8 = 2240 $$