FINDING THE SPECIFIC TERM OF AN ARITHMETIC SEQUENCE WORKSHEET

Problem 1 :

The first term of an arithmetic sequence is 5 and the common difference is 7. Find the 36^th term.

Problem 2 :

The 5^th term and 12^th terms of an arithmetic sequence are 14 and 35 respectively. Find 56^th term.

Problem 3 :

The sum of n terms of an arithmetic sequence is 3n² + 5n. Find the 25^th term.

Problem 4 :

The sum of the first two terms of an arithmetic series is 13 and the sum of the first four terms is 46. Determine the 15^th term.

Problem 5 :

A person is employed in a company at $3000 per month and he would get an increase of $100 per year. Find the monthly salary in 24^th year.

Problem 6 :

The m^th term of an arithmetic sequence is m and n^th term is m. Find the r^th term.

Answers

1. Answer :

Formula to find n^th term of an arithmetic sequence :

t_n = t₁ + (n - 1)d

Substitute n = 36, t₁ = 5 and d = 7.

t₃₆ = 5 + (36 - 1)(7)

= 5 + (35)(7)

= 5 + 245

= 250

36^th term is 250.

2. Answer :

Formula to find n^th term of an arithmetic sequence :

t_n = t₁ + (n - 1)d

Solve (1) and (2) for t₁ and d.

(2) - (1) :

7d = 21

Divide both sides by 7.

d = 3

Substitute d = 3 in (1).

t₁ + 4(3) = 14

t₁ + 12 = 14

Subtract 12 from both sides.

t₁ = 2

t_n = t₁ + (n - 1)d

Substitute n = 56, t₁ = 2 and d = 3.

t₅₆ = 2 + (56 - 1)(3)

= 2 + (55)(3)

= 2 + 165

= 167

56^th term is 167.

3. Answer :

S_n = 3n² + 5n

S₁ = 8 means sum of 1 term of the arithmetic sequence.

Sum of 1 term of an arithmetic sequence is the first term of the sequence.

t₁ = 8

S₂ = 22 means sum of the first two terms of the arithmetic sequence.

S₂ = 22

t₁ + t₂ = 22

Substitute t₁ = 8.

8 + t₂ = 22

Subtract 8 from both sides.

t₂ = 14

t₁ + (2 - 1)d = 14

t₁ + d = 14

Substitute t₁ = 8.

8 + d = 14

Subtract 8 from both sides.

d = 6

t_n = t₁ + (n - 1)d

Substitute n = 25, t₁ = 8 and d = 6.

t₂₅ = 8 + (25 - 1)(6)

= 8 + (24)(6)

= 8 + 144

= 152

25^th term is 152.

4. Answer :

Formula to find sum of first n terms of an arithmetic sequence :

S_n = (n/2)[2t₁ + (n - 1)d]

(2) - (1) :

2d = 10

Divide both sides by 2.

d = 5

Substitute d = 5 in (1).

2t₁ + 5 = 13

Subtract 5 from both sides.

2t₁ = 8

Divide both sides by 2.

t₁ = 4

t_n = t₁ + (n - 1)d

Substitute n = 15, t₁ = 4 and d = 5.

t₁₅ = 4 + (15 - 1)(5)

= 4 + (14)(5)

= 4 + 70

= 74

15^th term is 74.

5. Answer :

Monthly salary in 1^st year = $3000

Monthly salary in 2^nd year = $3100

Monthly salary in 3^rd year = $3200

3000, 3100, 3200, ..........

The above sequence is an arithmetic sequence with the first term 3000 and common difference 100.

t₁ = 3000 and d = 100

t_n = t₁ + (n - 1)d

Substitute n = 24, t₁ = 3000 and d = 100.

t₂₄ = 3000 + (24 - 1)(100)

= 3000 + (23)(100)

= 3000 + 2300

= 5300

The monthly salary in 24^th year is $5300.

6. Answer :

t_n = t₁ + (n - 1)d

(1) - (2) :

(m - 1)d - (n - 1)d = n - m

d[(m - 1) - (n - 1)] = n - m

d(m - 1 - n + 1) = n - m

d(m - n) = n - m

d(m - n) = -(m - n)

Divide both sides by (m - n).

d = -1

Substitute d = -1 in (1).

t₁ + (m - 1)(-1) = n

t₁ - m + 1 = n

t₁ = m + n - 1

t_n = t₁ + (n - 1)d

Substitute n = r, t₁ = m + n - 1 and d = -1.

t_r = (m + n - 1) + (r - 1)(-1)

= m + n - 1 - r + 1

= m + n - r

r^h term is (m + n - r).

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