# Arithmetic Sequences Problems with Solutions

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

an = a1 + (n - 1 )d

The sum s
n of the first n terms of an arithmetic sequence is defined by

sn = a1 + a2 + a3 + ... + a n

and is a
1 is given by

sn = n (a1 + an) / 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 a1 = 6 in the formula for the n th term given above

an = a1 + (n - 1 )d

= 6 + 3 (n - 1)

= 3 n + 3

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

a50 = 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

 10 S = ∑ (2n + 1 / 2) n=1

Solution to Problem 8:

• Let us first decompose this sum as follows

 10 S = ∑ (2n + 1 / 2) n=1
 10 10 = 2 ∑ n + ∑ (1 / 2) i=1 n=1

• 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 a3 = 9 and a8 = 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 a20 = 466

d) Find s
30 given that a10 = 28 and a20 = 58

e) Find the sum S defined by
 20 S = ∑ (3n - 1 / 2) n=1

f) Find the sum S defined by
 20 40 S = ∑ 0.2 n + ∑ 0.4 j n=1 j=21
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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