Arithmetic Sequence Calculator

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**Step 1 :**

**Step 2 :**

Arithmetic Sequence Calculator is an online tool used to find n^{th} term and sum of the sequence.

Arithmetic sequence is a sequence of numbers such that the distinction between two consecutive members of the sequence as the same common difference.

Arithmetic sequence is a sequence of numbers such that the distinction between two consecutive members of the sequence as the same common difference.

The formula for the n^{th} term an of an infinite arithmetic sequence with a common difference d and a first term a_{1} is given by

a_{n} = a_{1} + (n - 1 )d

The sum s_{n} of the first n terms of an infinite arithmetic sequence is defined by

s_{n} = a_{1} + a_{2} + a_{3} + ... + a_{n}

and is a1 is given by

s_{n} = n (a_{1} + a_{n}) / 2

Where

a_{1} = first number of the sequence

d = common difference between the sequence

a_{n} = nth term of the sequence

n = total number of value in the sequence.

Sn = Sum of nth term of the sequence

Put the values in the formulas and calculate it further.

Find the sum of all positive integers, from 27 to 4590 inclusive, which are divisible by 27.

**Step 1 :**Given: Sequence of first few numbers of positive integers divisible by 27 are given by 27, 54, 81...

**First term of the sequence 'a' = 27**

**Common difference 'd' = 27.**

We need to know the rank of the term 4590.

We use the following formula for the n^{th}term

a_{n}= a_{1}+ (n - 1 )d

The sum s

_{n}of the first n terms of an infinite arithmetic sequence is defined by

s_{n}= a_{1}+ a_{2}+ a_{3}+ ... + a_{n}

and is a1 is given by

s_{n}= n (a_{1}+ a_{n}) / 2**Step 2 :**Put the values in the formula and solve it further.

a

_{n}= a_{1}+ (n - 1 )d

4590 = a_{1}+ (n - 1 )d

Substitute a_{1}and d by their values

4590 = 27 + 27(n - 1)

Solve for n to obtain

**n = 170**

4590 is the 170^{th}term, we can use the following formula to find sum

s_{n}= n (a_{1}+ a_{n}) / 2

s_{170}= 170 (27 + 4590) / 2 = 392445.**Answer :****Term at position 170**= 4590**Sum of all terms till position**= 392445.Find the sum of all positive integers, from 23 to 3680 inclusive, which are divisible by 23.

**Step 1 :**Given: Sequence of first few numbers of positive integers divisible by 23 are given by 23, 46, 69...

**First term of the sequence 'a' = 23**

**Common difference 'd' = 23**

We need to know the rank of the term 3680.

We use the following formula for the n^{th}term

a_{n}= a_{1}+ (n - 1 )d

The sum s

_{n}of the first n terms of an infinite arithmetic sequence is defined by

s_{n}= a_{1}+ a_{2}+ a_{3}+ ... + a_{n}

and is a1 is given by

s

_{n}= n (a_{1}+ a_{n}) / 2**Step 2 :**Put the values in the formula and solve it further.

a

_{n}= a_{1}+ (n - 1 )d

3680 = a

_{1}+ (n - 1 )d

Substitute a_{1}and d by their values

3680 = 23 + 23(n - 1)

Solve for n to obtain

**n = 160**

3680 is the 160^{th}term, we can use the following formula to find sum

s_{n}= n (a_{1}+ a_{n}) / 2

s_{160}= 160 (23 + 3680) / 2 = 296240.**Answer :****Term at position 170**= 3680**Sum of all terms till position**= 296240.