Arithmetic sequences examples

Example: Find the sum of all natural numbers greater than 30 and smaller than 330 whose last digit is 1.

Solution: Given numbers form the sequence 31, 41, 51, . . . , 291, 301, 311, 321.

Therefore, a₁ = 31, a_n = 321 and d = 10

as a_n = a₁ + (n - 1) · d

then 321 = 31 + (n - 1) · 10 S_n = (n/2) · (a₁ + a_n)

290 = (n - 1) · 10 S₃₀ = 30/2 · (31 + 321)

n - 1 = 29, n = 30. S₃₀ = 15 · 352 = 5280.

Example: Find the first term and number of terms of an arithmetic sequence with a_n = 15, S_n = 64 and common difference d = 2.

Solution: Use the formulas for a_n and S_n to form a system of two equations in two unknowns,

(1) a_n = a₁ + (n - 1) · d

(2) S_n= (n/2) · [2a₁ + (n - 1) · d]

(1) a₁ + (n - 1) · 2 = 15, => a₁ = 17 - 2n

(2) (n/2) · [2a₁ + (n - 1) · 2] = 64, => n · (a₁+ n - 1) = 64

(1) a₁ = 17 - 2n => (2) n² + a₁n - n = 64

n² + (17 - 2n) · n - n = 64,

n² - 16n + 64 = 0,

n_1,2 = 8 ± Ö 64 - 64, n = 8, a₁ = 17 - 2 · 8 = 1.

Therefore, the sequence is 1, 3, 5, 7, 9, 11, 13, 15, . . . , where a₈ = 15 and S₈ = 64.

Example: Which arithmetic sequence has the property that the sum of its first five terms is 35 and the sum of the second and the sixth term is 20.

Solution: Use the formulas for S_n and a_nto form a system of two equations,

S_n= (n/2) · [2a₁ + (n - 1) · d] => (1) (5/2) · [2a₁ + 4d] = 35, a₁ + 2d = 7

a_n = a₁ + (n - 1) · d, a₂ + a₆ = 20 => (2) (a₁ + d) + (a₁ + 5d) = 20, 2a₁ + 6d = 20

(2) a₁ + 3d = 10

(2) - (1) (a₁ + 3d) - (a₁ + 2d) = 10 - 7, d = 3

(2) a₁ + 3d = 10 => a₁ = 10 - 3d = 10 - 3 · 3 = 1, a₁ = 1.

Thus, the sequence is 1, 4, 7, 10, 13, 16, 19, . . .

Example: How many natural numbers, divisible by 7, lie between 0 and 100, and what is their sum.

Solution: The sequence of numbers divisible by 7 is; 7, 14, 21, 28, . . . , 84, 91, 98

and since a_n = a₁ + (n - 1) · d, then 98 = 7 + (n - 1) · 7 | ¸ 7

S_n = n/2 · (a₁ + a_n), n = 14,

S₁₄ = 14/2 · (7 + 98) = 7 · 105 = 735.

Example: Find n and S_n of an arithmetic sequence if given are; a₁, a_n, and d.

Solution: Since, a_n = a₁ + (n - 1) · d then a_n - a₁ = (n - 1) · d, or n - 1 = (a_n - a₁) / d

n = (a_n - a₁ + d) / d

and S_n = (n/2) · (a₁ + a_n)

by substituting n = (a_n - a₁ + d) / d obtained is S_n = (a_n - a₁ + d) · (a₁ + a_n) / (2d)

Example: Find d and n of an arithmetic sequence if given are; a₁, a_n, and S_n.

Solution: Since, a_n = a₁ + (n - 1) · d and S_n = (n/2) · (a₁ + a_n)

d = (a_n - a₁) / (n - 1) <= n = (2S_n) / (a₁ + a_n)

d = (a_n - a₁) / [(2S_n) / (a₁ + a_n) - 1] = (a_n² - a₁²) / (2S_n - a₁ - a_n)

Example: Derive the formula for the nth odd natural number and the sum of the first n odd natural numbers.

Solution: The sequence of odd natural numbers is 1, 3, 5, 7, . . . , a_n, . . .

using a_n = a₁ + (n - 1) · d

a_n = 1 + (n - 1) · 2 and since S_n = (n/2) · (a₁ + a_n)

a_n = 2n - 1 => S_n then S_n = (n/2) · [1 + (2n - 1)] so, S_n = n².

Example: In a sequence of natural numbers divisible by 7 find the one that equals the one seventeenth of the sum of all natural numbers that precede this one.

Solution: A natural number divisible by 7 is 7n, where n Î N, so the sequence of the numbers is

7, 14, 21, 28, 35, 42, 49, . . .

and since the formula for the sum of the first n natural numbers S_n= n · (n + 1)/2, and the sum of natural numbers that precede the number divisible by 7 is, (7n + 1)[(7n - 1) + 1] / 2 then, the following equation meets the given condition

7n = (1/17) · 7n(7n - 1) / 2 thus, 7n = 35.

Pre-calculus contents K