# Arithmetic Sequences Problems with Solutions

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 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
\[
a_n = a_1 + (n - 1) d
\]
The sum s_{n} of the first n terms of an arithmetic sequence is defined by
\[
s_n = a_1 + a_2 + a_3 + ... + a_n
\]
and is is given by
\[
s_n = \dfrac{n (a_1 + a_n)}{2}
\]
Arithmetic Series Online Calculator. An online calculator to calculate the sum of the terms in an arithmetic sequence.

## Problems with Solutions

__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_{1} = 6 in the formula for the n th term given above

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

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

\[
a_{50} = 3 (50) + 3 = 153
\]

__Problem 2__

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

__Solution to Problem 2:__

Use the value of the common difference d = -10 and the first term a_{1} = 200 in the formula for the n th term given above and then apply it to the 20 th term

a_{20} = 200 + (-10) (20 - 1 ) = 10

__Problem 3__

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

__Solution to Problem 3:__

We use the n th term formula for the 6 th term, which is known, to write

a_{6} = 52 = a_{1} + 10 (6 - 1 )

The above equation allows us to calculate a_{1}.

a_{1} = 2

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_{15} = 2 + 10 (15 - 1) = 142

__Problem 4__

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.

__Solution to Problem 4:__

We use the n th term formula for the 5 th and 15 th terms to write

a_{5} = a_{1} + (5 - 1 ) d = 22

a_{15} = a_{1} + (15 - 1 ) d = 62

We obtain a system of 2 linear equations where the unknown are a_{1} and d. Subtract the right and left term of the two equations to obtain

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_{1}.

a_{1} + (5 - 1 ) 4 = 22

Solve for a_{1} to obtain.

a_{1} = 6

Now that we have calculated a_{1} and d we use them in the n th term formula to find the 100 th formula.

a_{100} = 6 + 4 (100 - 1 )= 402

__Problem 5__

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

__Solution to Problem 5:__

The sequence of integers starting from 1 to 1000 is given by

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_{1000} = 1000 (1 + 1000) / 2 = 500500

__Problem 6__

Find the sum of the first 50 even positive integers.

__Solution to Problem 6:__

The sequence of the first 50 even positive integers is given by

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_{50} = 2 + 2 (50 - 1) = 100

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_{50} = 50 (2 + 100) / 2 = 2550

__Problem 7__

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

__Solution to Problem 7:__

The first few terms of a sequence of positive integers divisible by 5 is given by

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_{1} + (n - 1 )d

Substitute a_{1} and d by their values

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_{311} = 311 (5 + 1555) / 2 = 242580

__Problem 8__

Find the sum S defined by
\[
S = \sum_{n=1}^{10} (2n + 1 / 2)
\]

__Solution to Problem 8:__

Let us first decompose this sum as follows

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

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

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_{n} = n (a_{1} + a_{n}) / 2 as follows

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

## Exercises

Answer the following questions related to arithmetic sequences:

a) Find a_{20} given that a_{3} = 9 and a_{8} = 24

b) Find a_{30} given that the first few terms of an arithmetic sequence are given by 6,12,18,...

c) Find d given that a_{1} = 10 and a_{20} = 466

d) Find s_{30} given that a_{10} = 28 and a_{20} = 58

e) Find the sum S defined by
\[
S = \sum_{n=1}^{20}(3n - 1 / 2)
\]

f) Find the sum S defined by
\[
S = \sum_{n=1}^{20}0.2 n + \sum_{j=21}^{40} 0.4 j
\]

## Solutions to Above Exercises

a) a_{20} = 60

b) a_{30} = 180

c) d = 24

d) s_{30} = 1335

e) 1380

f) 286

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