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 a_n 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 \(s_n\) 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

Problem1

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.
\[ a_{50} = 3 (50) + 3 = 153 \]

Problem2

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 \[ a_{20} = 200 + (-10) (20 - 1 ) = 10 \]

Problem3

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. \[ a_{15} = 2 + 10 (15 - 1) = 142 \]

Problem4

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 \( a_1 \) 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. \[ a_{100} = 6 + 4 (100 - 1 )= 402 \]

Problem5

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 is an arithmetic sequence with 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). \[ s_{1000} = 1000 (1 + 1000) / 2 = 500500 \]

Problem6

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 \[ s_{50} = 50 (2 + 100) / 2 = 2550 \]

Problem7

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 \(a_1\) 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 \[ s_{331} = 311 (5 + 1555) / 2 = 242580 \]

Problem8

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 \( \sum_{n=1}^{10} n \) 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 \( \sum_{n=1}^{10} (1/2) \) 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:

Find \( a_{20} \) given that \( a_3 = 9 \) and \( a_8 = 24 \).
Find \( a_{30} \) given that the first few terms of an arithmetic sequence are given by \( 6,12,18,...\)
Find d given that \( a_1 = 10 \) and \( a_{20} = 466 \)
Find \( s_{30} \) given that \( a_{10} = 28 \) and \( a_{20} = 58 \)
Find the sum S defined by \[ S = \sum_{n=1}^{20}(3n - 1 / 2) \]
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

\( a_{20} = 60 \)
\( a_{30} = 180 \)
\( d = 24 \)
\( s_{30} = 1335 \)
1380
286