Complete Math Magic ✨

All 12 Arithmetic Progression Problems with Colorful Solutions!

1(i)

Find the sum of 3, 7, 11,… up to 40 terms.

AP Sum Formula:

Sₙ = n/2 × [2a + (n-1)d]

Step 1: Identify the first term (a) and common difference (d)
a = 3, d = 7 - 3 = 4

Step 2: Plug values into the formula
S₄₀ = 40/2 × [2×3 + (40-1)×4]

Step 3: Calculate inside brackets
= 20 × [6 + 39×4] = 20 × [6 + 156] = 20 × 162

Step 4: Final calculation
= 3240

1(ii)

Find the sum of 102, 97, 92,… up to 27 terms.

AP Sum Formula:

Sₙ = n/2 × [2a + (n-1)d]

Step 1: Identify a and d
a = 102, d = 97 - 102 = -5

Step 2: Plug into formula
S₂₇ = 27/2 × [2×102 + (27-1)×(-5)]

Step 3: Calculate inside brackets
= 13.5 × [204 + 26×(-5)] = 13.5 × [204 - 130] = 13.5 × 74

Step 4: Final calculation
= 999

1(iii)

Find the sum of 6 + 13 + 20 + ... + 97

AP Formulas:

aₙ = a + (n-1)d
Sₙ = n/2 × (a + aₙ)

Step 1: Identify a, d, and aₙ
a = 6, d = 13 - 6 = 7, aₙ = 97

Step 2: Find number of terms (n)
97 = 6 + (n-1)×7 → 91 = (n-1)×7 → n-1 = 13 → n = 14

Step 3: Calculate sum
S₁₄ = 14/2 × (6 + 97) = 7 × 103 = 721

How many consecutive odd integers beginning with 5 will sum to 480?

AP Sum Formula:

Sₙ = n/2 × [2a + (n-1)d]

Step 1: Identify known values
a = 5, d = 2 (odd numbers), Sₙ = 480

Step 2: Plug into formula
480 = n/2 × [2×5 + (n-1)×2]

Step 3: Simplify equation
480 = n/2 × [10 + 2n - 2] = n/2 × (2n + 8) = n(n + 4)

Step 4: Solve quadratic equation
n² + 4n - 480 = 0
Using quadratic formula: n = [-4 ± √(16 + 1920)]/2 = [-4 ± √1936]/2 = [-4 ± 44]/2
Positive solution: n = 40/2 = 20

Find the sum of first 28 terms of an A.P. whose nth term is 4n - 3.

AP Formulas:

aₙ = 4n - 3 (given)
Sₙ = n/2 × (a₁ + aₙ)

Step 1: Find first term (a₁)
a₁ = 4(1) - 3 = 1

Step 2: Find 28th term (a₂₈)
a₂₈ = 4(28) - 3 = 112 - 3 = 109

Step 3: Calculate sum
S₂₈ = 28/2 × (1 + 109) = 14 × 110 = 1540

The sum of first n terms of a certain series is given as 2n² - 3n. Show that the series is an A.P.

Key Concept:

To prove a series is AP, show that the difference between consecutive terms is constant

Step 1: Find nth term (aₙ)
aₙ = Sₙ - Sₙ₋₁ = [2n² - 3n] - [2(n-1)² - 3(n-1)]
= 2n² - 3n - [2(n² - 2n + 1) - 3n + 3]
= 2n² - 3n - 2n² + 4n - 2 + 3n - 3
= (2n² - 2n²) + (-3n + 4n + 3n) + (-2 - 3)
= 4n - 5

Step 2: Find common difference (d)
aₙ₊₁ - aₙ = [4(n+1) - 5] - [4n - 5] = 4n + 4 - 5 - 4n + 5 = 4
Since difference is constant (4), the series is AP

The 104th term and 4th term of an A.P. are 125 and 0. Find the sum of first 35 terms.

AP Term Formula:

aₙ = a + (n-1)d

Step 1: Set up equations for given terms
a₄ = a + 3d = 0 → a = -3d
a₁₀₄ = a + 103d = 125

Step 2: Substitute a from first equation into second
-3d + 103d = 125 → 100d = 125 → d = 1.25

Step 3: Find a
a = -3 × 1.25 = -3.75

Step 4: Calculate sum of first 35 terms
S₃₅ = 35/2 × [2×(-3.75) + (35-1)×1.25]
= 17.5 × [-7.5 + 42.5] = 17.5 × 35 = 612.5

Find the sum of all odd positive integers less than 450.

AP Formulas:

Sequence: 1, 3, 5,..., 449
Sₙ = n/2 × (a + aₙ)

Step 1: Identify values
a = 1, d = 2, aₙ = 449

Step 2: Find number of terms
449 = 1 + (n-1)×2 → 448 = (n-1)×2 → n-1 = 224 → n = 225

Step 3: Calculate sum
S₂₂₅ = 225/2 × (1 + 449) = 112.5 × 450 = 50625

Find the sum of all natural numbers between 602 and 902 which are not divisible by 4.

Strategy:

Sum all numbers between 602-902 minus sum of numbers divisible by 4 in that range

Step 1: Sum all numbers from 603 to 901
This is AP with a=603, l=901, n=901-603+1=299
S₁ = 299/2 × (603 + 901) = 149.5 × 1504 = 224,848

Step 2: Sum numbers divisible by 4 from 604 to 900
First term (a) = 604, last (l) = 900, d = 4
Number of terms: n = (900-604)/4 + 1 = 296/4 + 1 = 75
S₂ = 75/2 × (604 + 900) = 37.5 × 1504 = 56,400

Step 3: Subtract to get final sum
Required sum = S₁ - S₂ = 224,848 - 56,400 = 168,448

Raghu wish to buy a laptop. He can buy it by paying ₹40,000 cash or by giving it in 10 installments as ₹4800 in the first month, ₹4750 in the second month, ₹4700 in the third month and so on. If he pays the money in this fashion, find (i) total amount paid in 10 installments (ii) how much extra amount that he has to pay than the cost?

AP Formulas:

Installments form AP: 4800, 4750, 4700,... (10 terms)
Sₙ = n/2 × [2a + (n-1)d]

Part (i): Total amount paid
a = 4800, d = -50, n = 10
S₁₀ = 10/2 × [2×4800 + (10-1)×(-50)]
= 5 × [9600 - 450] = 5 × 9150 = ₹45,750

Part (ii): Extra amount paid
Cash price = ₹40,000
Extra amount = ₹45,750 - ₹40,000 = ₹5,750

A man repays a loan of ₹65,000 by paying ₹400 in the first month and then increasing the payment by ₹300 every month. How long will it take for him to clear the loan?

AP Sum Formula:

Payments form AP: 400, 700, 1000,...
Sₙ = n/2 × [2a + (n-1)d] = 65,000

Step 1: Identify values
a = 400, d = 300, Sₙ = 65,000

Step 2: Plug into formula
65,000 = n/2 × [2×400 + (n-1)×300]
65,000 = n/2 × [800 + 300n - 300]
65,000 = n/2 × (500 + 300n)
65,000 = 250n + 150n²

Step 3: Solve quadratic equation
150n² + 250n - 65,000 = 0
Divide by 50: 3n² + 5n - 1300 = 0
Using quadratic formula:
n = [-5 ± √(25 + 15600)]/6 = [-5 ± √15625]/6 = [-5 ± 125]/6
Positive solution: n = 120/6 = 20 months

A brick staircase has a total of 30 steps. The bottom step requires 100 bricks. Each successive step requires two bricks less than the previous step. (i) How many bricks are required for the top most step? (ii) How many bricks are required to build the stair case?

AP Formulas:

Bricks per step: 100, 98, 96,... (30 terms)
aₙ = a + (n-1)d
Sₙ = n/2 × (a + aₙ)

Part (i): Bricks in top step
a = 100, d = -2, n = 30
a₃₀ = 100 + (30-1)×(-2) = 100 - 58 = 42 bricks

Part (ii): Total bricks
S₃₀ = 30/2 × (100 + 42) = 15 × 142 = 2130 bricks

If S₁, S₂, S₃,..., Sₘ are the sums of n terms of m A.P.'s whose first terms are 1,2,3,...,m and whose common differences are 1,3,5,...,(2m-1) respectively, then show that S₁ + S₂ + S₃ + ... + Sₘ = ½mn(mn + 1).

AP Sum Formula:

For k-th AP: Sₖ = n/2 × [2aₖ + (n-1)dₖ]

Step 1: Write general term for Sₖ
Given aₖ = k, dₖ = (2k-1)
Sₖ = n/2 × [2k + (n-1)(2k-1)]

Step 2: Expand Sₖ
= n/2 × [2k + (n-1)(2k) - (n-1)]
= n/2 × [2k + 2k(n-1) - (n-1)]
= n/2 × [2k(1 + n - 1) - (n-1)]
= n/2 × [2kn - (n-1)]

Step 3: Sum all Sₖ from k=1 to m
ΣSₖ = n/2 × [2nΣk - Σ(n-1)] from k=1 to m
= n/2 × [2n × m(m+1)/2 - m(n-1)]
= n/2 × [nm(m+1) - m(n-1)]
= nm/2 × [n(m+1) - (n-1)]
= nm/2 × [nm + n - n + 1]
= nm/2 × (nm + 1)
= ½mn(mn + 1)

Find the sum of (a-b)/(a+b) + (3a-2b)/(a+b) + (5a-3b)/(a+b) + ... to 12 terms.

AP Sum Formula:

Series can be separated into two APs with common denominator

Step 1: Separate numerator
Series = [a/(a+b) + 3a/(a+b) + 5a/(a+b) + ...] - [b/(a+b) + 2b/(a+b) + 3b/(a+b) + ...]

Step 2: First AP (a terms):
a/(a+b) × [1 + 3 + 5 + ... 12 terms]
Sum = a/(a+b) × n/2 × [2a₁ + (n-1)d] = a/(a+b) × 12/2 × [2×1 + 11×2] = 6a/(a+b) × 24 = 144a/(a+b)

Step 3: Second AP (b terms):
b/(a+b) × [1 + 2 + 3 + ... 12 terms]
Sum = b/(a+b) × n(n+1)/2 = b/(a+b) × 12×13/2 = 78b/(a+b)

Step 4: Combine results
Total sum = 144a/(a+b) - 78b/(a+b) = (144a - 78b)/(a+b)
= 6(24a - 13b)/(a+b)