# 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 an of an arithmetic sequence with a common difference d and a first term a1 is given by $a_n = a_1 + (n - 1) d$ The sum sn 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.

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 a1 = 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 a1 = 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 a1.
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 a1 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 a1.
a
1 + (5 - 1 ) 4 = 22
Solve for a1 to obtain.
a
1 = 6
Now that we have calculated a1 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 a1 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 sn = n (a1 + an) / 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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math problems with detailed solutions in this site.