Question

How do you find the sum of the arithmetic sequence $2 + 5 + 8 + ..... + 56$?

Answer 1

Here, we will first find the common difference between the two terms of the given series. Then we will substitute all the values in the formula of the ${n^{th}}$ term of arithmetic progression to get the number of terms in the sequence. Finally, we will substitute all the values in the formula of the sum of ${n^{th}}$ terms of an arithmetic sequence to get the desired answer.

Formula used:
We will use the following formulas:
${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$, where $n =$ number of terms,$a =$ First term,$d =$ Common difference.
${n^{th}}$ term formula: $a + \left( {n - 1} \right)d =$ Last term, where, $n =$ number of terms,$a =$ First term,$d =$ Common difference.

Complete step by step solution:
The sequence given to us is: $2 + 5 + 8 + ..... + 56$
Here, the value of the first term is $a = 2$.
Now we will find the common difference, $d$.
$d = 5 - 2 = 8 - 5 = 3$
Also, the last term is 56.
Now we have to find the number of terms by using the ${n^{th}}$ term formula.
Substituting the values of first term, last term and common difference in the formula $a + \left( {n - 1} \right)d =$ Last term, we get
$2 + \left( {n - 1} \right) \times 3 = 56$
Multiplying the terms, we get
$\Rightarrow 2 + 3n - 3 = 56$
Rewriting the equation, we get
$\Rightarrow 3n = 56 + 3 - 2$
Adding and subtracting the like terms, we get
$\Rightarrow n = \dfrac{{57}}{3} = 19$
Now we will find the sum of the given sequence.
Substitute $a = 2,d = 3$ and $n = 19$ in the formula ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$, we get
${S_{19}} = \dfrac{{19}}{2}\left( {2 + \left( {19 - 1} \right) \times 3} \right)$
Simplifying the expression, we get
$\Rightarrow {S_{19}} = \dfrac{{19}}{2}\left( {2 + 54} \right)$
$\Rightarrow {S_{19}} = 532$

Therefore, the sum of the arithmetic sequence is 532.

Note:
We can find the sum of the arithmetic sequence by adding the first and last term and then divide the sum by two.
Now we know that,
First term $= 2$
Last term $= 56$
Common difference $= 5 - 2 = 3$
${n^{th}}$ term formula: $a + \left( {n - 1} \right)d =$ Last term
$2 + \left( {n - 1} \right) \times 3 = 56$
Multiplying the terms, we get
$\Rightarrow 2 + 3n - 3 = 56$
Rewriting the equation, we get
$\Rightarrow 3n = 56 + 3 - 2$
Adding and subtracting the like terms, we get
$\Rightarrow n = \dfrac{{57}}{3} = 19$
Now average of first and last term $= \dfrac{{2 + 56}}{2} = \dfrac{{58}}{2} = 29$
Now we will find the sum of the sequence by multiplying the average by the number of terms in the sequence.
Sum of sequence $=$ Number of terms $\times$ Average of first and last term
Substituting the values, we get
$\Rightarrow$ Sum of Sequence $= 19 \times 29 = 532$