Arithmetic Progression Problems

[MathDefinition title="Arithmetic Progression (AP)"] An arithmetic progression is a sequence where the difference between consecutive terms is constant. This constant difference is called the common difference (d). [/MathDefinition]

[MathDefinition title="nth Term of AP"] The nth term of an AP is given by: a_n = a + (n-1)d where a is the first term and d is the common difference. [/MathDefinition]

[MathKeyFormula title="Key Formulas for AP"]

1. nth term: a_n = a + (n-1)d
2. Sum of n terms: S_n = n/2[2a + (n-1)d] = n/2(a + a_n)
3. Common difference: d = a_n - a_{n-1}
4. Number of terms: n = (l - a)/d + 1, where l is the last term
5. Arithmetic mean of a and b: (a + b)/2 [/MathKeyFormula]

[MathNote title="Problem Solving Strategy"]

1. Identify what is given (a, d, n, a_n, or S_n)
2. Choose the appropriate formula
3. Substitute values and solve
4. Check if the answer makes sense [/MathNote]

[MathExample title="Example 1: Finding nth Term"] Problem: Find the 20th term of the AP: 3, 7, 11, 15, ...

Solution: Here a = 3, d = 7 - 3 = 4, n = 20

Using a_n = a + (n-1)d: a_20 = 3 + (20-1) * 4 = 3 + 19 * 4 = 3 + 76 = 79

Therefore, the 20th term is 79. [/MathExample]

[MathExample title="Example 2: Finding Sum"] Problem: Find the sum of the first 30 terms of the AP: 5, 8, 11, 14, ...

Solution: Here a = 5, d = 8 - 5 = 3, n = 30

Using S_n = n/2[2a + (n-1)d]: S_30 = 30/2[2(5) + (30-1)(3)] = 15[10 + 87] = 15 * 97 = 1455

Therefore, the sum of the first 30 terms is 1455. [/MathExample]

[MathExample title="Example 3: Finding Number of Terms"] Problem: How many terms of the AP 1, 4, 7, 10, ... must be taken to give a sum of 715?

Solution: Here a = 1, d = 3, S_n = 715

Using S_n = n/2[2a + (n-1)d]: 715 = n/2[2(1) + (n-1)(3)] 715 = n/2[2 + 3n - 3] 715 = n/2[3n - 1] 1430 = 3n^2 - n 3n^2 - n - 1430 = 0

Using quadratic formula: n = [1 ± √(1 + 17160)]/6 = [1 ± √17161]/6 = [1 ± 131]/6

n = 132/6 = 22 or n = -130/6 (rejected)

Therefore, 22 terms must be taken. [/MathExample]

[MathExample title="Example 4: Word Problem"] Problem: The sum of three numbers in AP is 27 and their product is 450. Find the numbers.

Solution: Let the three numbers be a-d, a, a+d

Sum: (a-d) + a + (a+d) = 3a = 27, so a = 9

Product: (9-d)(9)(9+d) = 450 9(81 - d^2) = 450 81 - d^2 = 50 d^2 = 31 d = √31

The three numbers are: 9 - √31, 9, 9 + √31

Verification: Sum = 9 - √31 + 9 + 9 + √31 = 27 ✓ Product = 9(81 - 31) = 9 * 50 = 450 ✓ [/MathExample]