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 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.



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 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
a20 = 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
a6 = 52 = a1 + 10 (6 - 1 )
The above equation allows us to calculate a1.
a1 = 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.
a15 = 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
a5 = a1 + (5 - 1 ) d = 22
a15 = a1 + (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.
a1 + (5 - 1 ) 4 = 22
Solve for a1 to obtain.
a1 = 6
Now that we have calculated a1 and d we use them in the n th term formula to find the 100 th formula.
a100 = 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).
s1000 = 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
a50 = 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
s50 = 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 = a1 + (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
s311 = 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 a20 given that a3 = 9 and a8 = 24
b) Find a30 given that the first few terms of an arithmetic sequence are given by 6,12,18,...
c) Find d given that a1 = 10 and a20 = 466
d) Find s30 given that a10 = 28 and a20 = 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) a20 = 60
b) a30 = 180
c) d = 24
d) s30 = 1335
e) 1380
f) 286



More References and links

  1. Geometric Sequences Problems with Solutions
  2. math problems with detailed solutions
  3. Math Tutorials and Problems

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