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 sn of the first n terms of an arithmetic sequence is defined by

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


and is a1 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 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
      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) a20 = 60

b) a30 = 180

c) d = 24

d) s30 = 1335

e) 1380

f) 286
More math problems with detailed solutions in this site.

Arithmetic Series Online Calculator. An online calculator to calculate the sum of the terms in an arithmetic sequence.


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Updated: 2 April 2013

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