Arithmetic sequences are used throughout mathematics and applied to engineering, sciences, computer sciences, biology and finance problems.

A set of problems and exercises involving arithmetic sequences, along with detailed solutions and answers, are presented.

__REVIEW OF ARITHMETIC SEQUENCES__

The formula for the n th term a_{n} of an arithmetic sequence with a common difference d and a first term a_{1} is given by

a_n = a_1 + (n - 1) d

s_n = a_1 + a_2 + a_3 + ... + a_n

s_n = \dfrac{n (a_1 + a_n)}{2}

__Problem 1__

The first term of an arithmetic sequence is equal to 6 and the common difference is equal to 3. Find a formula for the n th term and the value of the 50 th term

** Solution to Problem 1:**
Use the value of the common difference d = 3 and the first term a

a_n = a_1 + (n - 1) d \\
= 6 + 3 (n - 1) \\
= 3 n + 3

The 50 th term is found by setting n = 50 in the above formula.

a_{50} = 3 (50) + 3 = 153

The first term of an arithmetic sequence is equal to 200 and the common difference is equal to -10. Find the value of the 20 th term

a

An arithmetic sequence has a common difference equal to 10 and its 6 th term is equal to 52. Find its 15 th term.

a

The above equation allows us to calculate a

a

Now that we know the first term and the common difference, we use the n th term formula to find the 15 th term as follows.

a

An arithmetic sequence has a its 5 th term equal to 22 and its 15 th term equal to 62. Find its 100 th term.

a

a

We obtain a system of 2 linear equations where the unknown are a

62 - 22 = 14 d - 4 d

Solve for d.

d = 4

Now use the value of d in one of the equations to find a

a

Solve for a

a

Now that we have calculated a

a

Find the sum of all the integers from 1 to 1000.

1 , 2 , 3 , 4 , ... , 1000

The above sequence has 1000 terms. The first term is 1 and the last term is 1000 and the common difference is equal to 1. We have the formula that gives the sum of the first n terms of an arithmetic sequence knowing the first and last term of the sequence and the number of terms (see formula above).

s

Find the sum of the first 50 even positive integers.

2 , 4 , 6 , ...

The above sequence has a first term equal to 2 and a common difference d = 2. We use the n th term formula to find the 50 th term

a

We now the first term and last term and the number of terms in the sequence, we now find the sum of the first 50 terms

s

Find the sum of all positive integers, from 5 to 1555 inclusive, that are divisible by 5.

5 , 10 , 15 , ...

The above sequence has a first term equal to 5 and a common difference d = 5. We need to know the rank of the term 1555. We use the formula for the n th term as follows

1555 = a

Substitute a

1555 = 5 + 5(n - 1 )

Solve for n to obtain

n = 311

We now know that 1555 is the 311 th term, we can use the formula for the sum as follows

s

Find the sum S defined by

S = \sum_{n=1}^{10} (2n + 1 / 2)

S = \sum_{n=1}^{10} (2n + 1 / 2)

= 2 \sum_{n=1}^{10} n + \sum_{n=1}^{10} (1/2)

The term ∑ n is the sum of the first 10 positive integers. The 10 first positive integers make an arirhmetic sequence with first term equal to 1, it has n = 10 terms and its 10 th term is equal to 10. This sum is obtained using the formula s

10(1+10)/2 = 55

The term ∑ (1 / 2) is the addition of a constant term 10 times and is given by

10(1/2) = 5

The sum S is given by

S = 2(55) + 5 = 115

Answer the following questions related to arithmetic sequences:

a) Find a

b) Find a

c) Find d given that a

d) Find s

e) Find the sum S defined by

S = \sum_{n=1}^{20}(3n - 1 / 2)

f) Find the sum S defined by

S = \sum_{n=1}^{20}0.2 n + \sum_{j=21}^{40} 0.4 j

a) a

b) a

c) d = 24

d) s

e) 1380

f) 286

More math problems with detailed solutions in this site.

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Updated: 28 July 2018 (A Dendane)