# 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 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

_{n}of the first n terms of an arithmetic sequence is defined by

__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

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

__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

**Use the value of the common difference d = -10 and the first term a**

__Solution to Problem 2:___{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.

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

__Solution to Problem 3:__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.

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

__Solution to Problem 4:__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.

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

__Solution to Problem 5:__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.

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

__Solution to Problem 6:__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.

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

__Solution to Problem 7:__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

**Let us first decompose this sum as follows**

__Solution to Problem 8:__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

f) Find the sum S defined by

**Solutions to Above Exercises:**

a) a

_{20}= 60

b) a

_{30}= 180

c) d = 24

d) s

_{30}= 1335

e) 1380

f) 286

More math problems with detailed solutions in this site.

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Updated: 28 July 2018 (A Dendane)