Question 1

The sum of an infinite geometric series with first term a = 6 and common ratio r = 1/2 is:

Accepted Answer

For an infinite geometric series with |r| < 1, the sum converges to:
S∞ = a / (1 - r)

Given: a = 6, r = 1/2

Step 1: Check convergence condition
|r| = |1/2| = 1/2 < 1 ✓ (series converges)

Step 2: Apply the infinite sum formula
S∞ = 6 / (1 - 1/2)

Step 3: Simplify the denominator
S∞ = 6 / (1/2)

Step 4: Divide by a fraction (multiply by reciprocal)
S∞ = 6 × 2 = 12

Verification: The series is 6 + 3 + 1.5 + 0.75 + ... Each term is half the previous. The partial sums approach 12 as n → ∞.

Question 2

The nth term of a sequence is given by aₙ = 2n² - 3n + 1. What is the sum of the first 5 terms?

Accepted Answer

Given: aₙ = 2n² - 3n + 1

Step 1: Calculate each of the first 5 terms
a₁ = 2(1)² - 3(1) + 1 = 2 - 3 + 1 = 0
a₂ = 2(2)² - 3(2) + 1 = 8 - 6 + 1 = 3
a₃ = 2(3)² - 3(3) + 1 = 18 - 9 + 1 = 10
a₄ = 2(4)² - 3(4) + 1 = 32 - 12 + 1 = 21
a₅ = 2(5)² - 3(5) + 1 = 50 - 15 + 1 = 36

Step 2: Sum the terms
S₅ = a₁ + a₂ + a₃ + a₄ + a₅ = 0 + 3 + 10 + 21 + 36 = 70

Alternative method using summation formulas:
S₅ = Σ(2n² - 3n + 1) from n=1 to 5
   = 2Σn² - 3Σn + Σ1
   = 2[1² + 2² + 3² + 4² + 5²] - 3[1 + 2 + 3 + 4 + 5] + 5
   = 2[1 + 4 + 9 + 16 + 25] - 3[15] + 5
   = 2(55) - 45 + 5
   = 110 - 45 + 5
   = 70 ✓

Question 3

In a geometric progression, the first term is 2 and the common ratio is 1/2. What is the sum of the first 6 terms?

Accepted Answer

For a geometric progression with first term a and common ratio r, the sum of the first n terms is:
Sₙ = a(1 - rⁿ)/(1 - r)  when r ≠ 1

Given: a = 2, r = 1/2, n = 6

S₆ = 2[1 - (1/2)⁶] / [1 - 1/2]
   = 2[1 - 1/64] / [1/2]
   = 2 × (63/64) / (1/2)
   = 2 × (63/64) × 2
   = 4 × (63/64)
   = 252/64
   = 63/16

Alternatively: 63/16 = 3.9375 or 3 15/16

Question 4

In an arithmetic progression, the 7th term is 28 and the 15th term is 60. What is the common difference?

Accepted Answer

For an arithmetic progression with first term a and common difference d:
aₙ = a + (n - 1)d

Given: a₇ = 28 and a₁₅ = 60

Step 1: Write equations using the AP formula
a₇ = a + 6d = 28 ... (1)
a₁₅ = a + 14d = 60 ... (2)

Step 2: Subtract equation (1) from equation (2)
(a + 14d) - (a + 6d) = 60 - 28
8d = 32
d = 4

Step 3: Verify by finding a
From equation (1): a + 6(4) = 28 → a = 28 - 24 = 4
Check: a₁₅ = 4 + 14(4) = 4 + 56 = 60 ✓

Question 5

The sum of an infinite geometric series with first term a = 5 and common ratio r = 1/3 is:

Accepted Answer

For an infinite geometric series with |r| < 1, the sum is given by:
S∞ = a / (1 - r)

Given: a = 5, r = 1/3

Step 1: Check convergence condition
|r| = |1/3| = 1/3 < 1 ✓ (series converges)

Step 2: Apply the infinite sum formula
S∞ = a / (1 - r)
S∞ = 5 / (1 - 1/3)
S∞ = 5 / (2/3)
S∞ = 5 × (3/2)
S∞ = 15/2
S∞ = 7.5

Step 3: Verify by checking the series
Series: 5 + 5(1/3) + 5(1/3)² + 5(1/3)³ + ...
= 5 + 5/3 + 5/9 + 5/27 + ...
This is a convergent geometric series with sum 15/2 = 7.5

Question 6

The sum of the series 2 + 4 + 8 + 16 + ... up to 10 terms is:

Accepted Answer

This is a geometric series with first term a = 2 and common ratio r = 2.

For a finite geometric series with n terms:
Sₙ = a(rⁿ - 1) / (r - 1)

Given: a = 2, r = 2, n = 10

Step 1: Identify the series type
Each term is double the previous: 2, 4, 8, 16, ...
This is a geometric series with a = 2 and r = 2.

Step 2: Apply the finite sum formula
S₁₀ = a(rⁿ - 1) / (r - 1)
S₁₀ = 2(2¹⁰ - 1) / (2 - 1)
S₁₀ = 2(1024 - 1) / 1
S₁₀ = 2(1023)
S₁₀ = 2046

Step 3: Verify by alternative formula
Sₙ = a(rⁿ - 1) / (r - 1) = 2(2¹⁰ - 1) = 2 × 1023 = 2046
Alternatively: S₁₀ = (2¹¹ - 2) / (2 - 1) = (2048 - 2) = 2046 ✓

Question 7

If the sum of the first n natural numbers is 55, then n equals:

Accepted Answer

The sum of the first n natural numbers is given by the formula:
Sₙ = n(n + 1) / 2

Given: Sₙ = 55

Step 1: Set up the equation
n(n + 1) / 2 = 55

Step 2: Multiply both sides by 2
n(n + 1) = 110

Step 3: Expand and rearrange
n² + n = 110
n² + n - 110 = 0

Step 4: Factor the quadratic
We need two numbers that multiply to -110 and add to 1.
Those numbers are 11 and -10.
(n + 11)(n - 10) = 0

Step 5: Solve for n
n = -11 or n = 10

Since n must be a positive integer (number of terms), n = 10.

Step 6: Verify
S₁₀ = 10(11) / 2 = 110 / 2 = 55 ✓
Alternatively: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55 ✓

Question 8

The 3rd term of an arithmetic progression is 7 and the 7th term is 19. What is the sum of the first 10 terms?

Accepted Answer

For an arithmetic progression with first term a and common difference d:
- nth term: aₙ = a + (n - 1)d
- Sum of first n terms: Sₙ = n/2 [2a + (n - 1)d] or Sₙ = n/2 (first term + last term)

Given: a₃ = 7 and a₇ = 19

From the nth term formula:
a₃ = a + 2d = 7  ... (1)
a₇ = a + 6d = 19 ... (2)

Subtracting (1) from (2):
4d = 12
d = 3

Substituting back into (1):
a + 2(3) = 7
a = 1

Now find S₁₀:
S₁₀ = 10/2 [2(1) + (10 - 1)(3)]
    = 5[2 + 27]
    = 5 × 29
    = 145

Question 9

The sum of an infinite geometric series with first term a and common ratio r is 8. If a = 2, what is the value of r?

Accepted Answer

For an infinite geometric series with |r| < 1, the sum is:
S∞ = a / (1 - r)

Given: S∞ = 8 and a = 2

Substitute into the formula:
8 = 2 / (1 - r)

Multiply both sides by (1 - r):
8(1 - r) = 2
8 - 8r = 2
-8r = -6
r = 6/8 = 3/4

Verification: Check that |r| < 1: |3/4| = 0.75 < 1 ✓
Check the sum: S∞ = 2 / (1 - 3/4) = 2 / (1/4) = 8 ✓

Question 10

In an arithmetic progression, the sum of the first n terms is Sₙ = 2n² + 3n. What is the common difference of the progression?

Accepted Answer

For an arithmetic progression, if Sₙ = 2n² + 3n, we can find the common difference by computing the first two terms.

Method 1: Using aₙ = Sₙ - Sₙ₋₁

For n = 1:
a₁ = S₁ = 2(1)² + 3(1) = 2 + 3 = 5

For n = 2:
S₂ = 2(2)² + 3(2) = 8 + 6 = 14
a₂ = S₂ - S₁ = 14 - 5 = 9

Common difference:
d = a₂ - a₁ = 9 - 5 = 4

Verification using the general formula for Sₙ in AP:
If Sₙ = An² + Bn, then d = 2A
Here A = 2, so d = 2(2) = 4 ✓

Alternative check: a₃ = S₃ - S₂ = [2(9) + 9] - 14 = 27 - 14 = 13
Indeed, 13 - 9 = 4 ✓

Question 11

The sum of the first n terms of a sequence is given by Sₙ = 3n² + 2n. What is the 5th term of the sequence?

Accepted Answer

To find the nth term of a sequence from the sum formula, we use: aₙ = Sₙ - Sₙ₋₁ for n ≥ 2.

Given: Sₙ = 3n² + 2n

Step 1: Find S₅ = 3(5)² + 2(5) = 3(25) + 10 = 75 + 10 = 85

Step 2: Find S₄ = 3(4)² + 2(4) = 3(16) + 8 = 48 + 8 = 56

Step 3: Apply the formula a₅ = S₅ - S₄ = 85 - 56 = 29

Verification: We can also check using a₁ = S₁ = 3(1)² + 2(1) = 5, and the sequence is 5, 11, 17, 23, 29, ... which is arithmetic with common difference d = 6.

Question 12

In a geometric progression, the first term is 2 and the common ratio is 3. What is the sum of the first 4 terms?

Accepted Answer

For a geometric progression with first term a and common ratio r, the sum of the first n terms is:
Sₙ = a(rⁿ - 1)/(r - 1) when r ≠ 1

Given: a = 2, r = 3, n = 4

Step 1: Apply the GP sum formula
S₄ = 2(3⁴ - 1)/(3 - 1)

Step 2: Calculate 3⁴ = 81
S₄ = 2(81 - 1)/2

Step 3: Simplify
S₄ = 2(80)/2 = 80

Verification: The terms are 2, 6, 18, 54. Sum = 2 + 6 + 18 + 54 = 80 ✓

Question 13

An arithmetic progression has first term a = 5 and common difference d = 3. If the sum of the first n terms is 440, find n.

Accepted Answer

For an arithmetic progression, the sum of the first n terms is:
Sₙ = n/2 [2a + (n-1)d]

Given: a = 5, d = 3, Sₙ = 440

Step 1: Substitute into the formula
440 = n/2 [2(5) + (n-1)(3)]

Step 2: Simplify inside the brackets
440 = n/2 [10 + 3n - 3]
440 = n/2 [7 + 3n]

Step 3: Multiply both sides by 2
880 = n(7 + 3n)
880 = 7n + 3n²

Step 4: Rearrange to standard form
3n² + 7n - 880 = 0

Step 5: Use the quadratic formula or factorization
Using the quadratic formula: n = [-7 ± √(49 + 4(3)(880))] / (2×3)
n = [-7 ± √(49 + 10560)] / 6
n = [-7 ± √10609] / 6
n = [-7 ± 103] / 6

Step 6: Solve for positive n
n = (-7 + 103)/6 = 96/6 = 16 (taking the positive root)

Verification: S₁₆ = 16/2 [2(5) + 15(3)] = 8[10 + 45] = 8(55) = 440 ✓

Question 14

The sum of the first n terms of a sequence is given by Sₙ = 3n² + 2n. Find the 5th term of the sequence.

Accepted Answer

To find the nth term of a sequence when the sum formula is given, we use: aₙ = Sₙ - Sₙ₋₁ for n ≥ 2.

Given: Sₙ = 3n² + 2n

For n = 5:
S₅ = 3(5)² + 2(5) = 3(25) + 10 = 75 + 10 = 85

For n = 4:
S₄ = 3(4)² + 2(4) = 3(16) + 8 = 48 + 8 = 56

Therefore: a₅ = S₅ - S₄ = 85 - 56 = 29

Verification using the general term formula:
For n ≥ 2: aₙ = Sₙ - Sₙ₋₁ = [3n² + 2n] - [3(n-1)² + 2(n-1)]
= 3n² + 2n - [3(n² - 2n + 1) + 2n - 2]
= 3n² + 2n - 3n² + 6n - 3 - 2n + 2
= 6n - 1

So a₅ = 6(5) - 1 = 30 - 1 = 29 ✓

Question 15

In a geometric progression, the 3rd term is 4 and the 6th term is 32. If all terms are positive, what is the sum of the first 5 terms?

Accepted Answer

In a GP with first term 'a' and common ratio 'r', the nth term is: aₙ = arⁿ⁻¹

Given:
- a₃ = 4, so ar² = 4 ... (1)
- a₆ = 32, so ar⁵ = 32 ... (2)

Dividing equation (2) by equation (1):
(ar⁵)/(ar²) = 32/4
r³ = 8
r = 2 (since all terms are positive)

Substituting r = 2 in equation (1):
a(2)² = 4
4a = 4
a = 1

The sum of first n terms of a GP is: Sₙ = a(rⁿ - 1)/(r - 1) when r ≠ 1

For n = 5:
S₅ = 1(2⁵ - 1)/(2 - 1)
S₅ = (32 - 1)/1
S₅ = 31

Verification: Terms are 1, 2, 4, 8, 16. Sum = 1 + 2 + 4 + 8 + 16 = 31 ✓

Question 16

The sum of an infinite geometric series with first term a and common ratio r is 6. If the first term is doubled and the common ratio is halved, the new sum is 8. Find the original common ratio r.

Accepted Answer

For an infinite GP with |r| < 1, the sum is: S = a/(1 - r)

Given condition 1:
a/(1 - r) = 6 ... (1)

Given condition 2 (first term doubled, common ratio halved):
2a/(1 - r/2) = 8 ... (2)

From equation (1): a = 6(1 - r) ... (3)

Substituting (3) into equation (2):
2[6(1 - r)]/(1 - r/2) = 8
12(1 - r)/(1 - r/2) = 8
12(1 - r) = 8(1 - r/2)
12 - 12r = 8 - 4r
12 - 8 = 12r - 4r
4 = 8r
r = 1/2

Verification:
From a = 6(1 - 1/2) = 6(1/2) = 3
Original sum: S = 3/(1 - 1/2) = 3/(1/2) = 6 ✓
New sum: S' = 2(3)/(1 - 1/4) = 6/(3/4) = 8 ✓

Question 17

The sum of the first n terms of a sequence is Sₙ = n(n + 1)(n + 2)/3. What is the sum of the 4th and 5th terms?

Accepted Answer

Given: Sₙ = n(n + 1)(n + 2)/3

To find individual terms, we use: aₙ = Sₙ - Sₙ₋₁ for n ≥ 2

For the 4th term:
S₄ = 4(5)(6)/3 = 120/3 = 40
S₃ = 3(4)(5)/3 = 60/3 = 20
a₄ = S₄ - S₃ = 40 - 20 = 20

For the 5th term:
S₅ = 5(6)(7)/3 = 210/3 = 70
S₄ = 4(5)(6)/3 = 120/3 = 40
a₅ = S₅ - S₄ = 70 - 40 = 30

Therefore: a₄ + a₅ = 20 + 30 = 50

Alternative approach (deriving general term):
aₙ = Sₙ - Sₙ₋₁ = n(n+1)(n+2)/3 - (n-1)n(n+1)/3
= [n(n+1)(n+2) - (n-1)n(n+1)]/3
= n(n+1)[(n+2) - (n-1)]/3
= n(n+1)(3)/3
= n(n+1)

So: a₄ = 4(5) = 20 and a₅ = 5(6) = 30
Sum = 50 ✓

Question 18

In an arithmetic progression, the sum of the first 10 terms is 100, and the sum of the first 20 terms is 300. Find the 15th term of the progression.

Accepted Answer

In an AP with first term 'a' and common difference 'd', the sum of first n terms is:
Sₙ = n/2 [2a + (n-1)d]

Given:
S₁₀ = 100, so 10/2 [2a + 9d] = 100
5[2a + 9d] = 100
2a + 9d = 20 ... (1)

S₂₀ = 300, so 20/2 [2a + 19d] = 300
10[2a + 19d] = 300
2a + 19d = 30 ... (2)

Subtracting equation (1) from equation (2):
(2a + 19d) - (2a + 9d) = 30 - 20
10d = 10
d = 1

Substituting d = 1 in equation (1):
2a + 9(1) = 20
2a + 9 = 20
2a = 11
a = 11/2 = 5.5

The nth term of an AP is: aₙ = a + (n-1)d

For the 15th term:
a₁₅ = 5.5 + (15-1)(1)
a₁₅ = 5.5 + 14
a₁₅ = 19.5

Verification:
S₁₀ = 10/2 [2(5.5) + 9(1)] = 5[11 + 9] = 5(20) = 100 ✓
S₂₀ = 20/2 [2(5.5) + 19(1)] = 10[11 + 19] = 10(30) = 300 ✓

Question 19

In a geometric progression, the 3rd term is 4 and the 6th term is 32. What is the sum of the first 5 terms of this GP?

Accepted Answer

In a GP with first term a and common ratio r:
aₙ = arⁿ⁻¹

Given:
a₃ = 4  ⟹  ar² = 4  ... (1)
a₆ = 32 ⟹  ar⁵ = 32 ... (2)

Dividing equation (2) by equation (1):
(ar⁵)/(ar²) = 32/4
r³ = 8
r = 2

Substituting r = 2 into equation (1):
a(2)² = 4
4a = 4
a = 1

Sum of first n terms of a GP: Sₙ = a(rⁿ - 1)/(r - 1) when r ≠ 1

For n = 5:
S₅ = 1(2⁵ - 1)/(2 - 1)
S₅ = (32 - 1)/1
S₅ = 31

Verification: The sequence is 1, 2, 4, 8, 16, 32, ...
Sum of first 5 terms: 1 + 2 + 4 + 8 + 16 = 31 ✓

Question 20

The sum of an infinite geometric series with first term a and common ratio r is 6. If the first term is 2, what is the common ratio?

Accepted Answer

For an infinite geometric series with |r| < 1, the sum is:
S∞ = a/(1 - r)

Given:
S∞ = 6
a = 2

Substituting into the formula:
6 = 2/(1 - r)

Multiplying both sides by (1 - r):
6(1 - r) = 2
6 - 6r = 2
-6r = 2 - 6
-6r = -4
r = -4/-6 = 2/3

Verification:
S∞ = 2/(1 - 2/3) = 2/(1/3) = 2 × 3 = 6 ✓

Also verify |r| < 1: |2/3| = 2/3 < 1 ✓ (convergence condition satisfied)

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