UPSKILL MATH PLUS

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In the previous chapter we explored about Arithmetic progression. In this chapter we learn about the Arithmetic series.

Arithmetic series is a series whose terms are in Arithmetic progression.

Consider the \(a, a +d, a +2d, a + 3d +\),... the Arithmetic Progression. The sum of first '\(n\)' terms of a Arithmetic Progression denoted by \(S_n\) is given by:

\(S_n =\) \(a + (a + d) + (a +2d) + (a + 3d)+... + (a + (n - 1)d)\)

**...…..….(1)**Let's rewriting the above in reverse order.

\(S_n =\) \( (a + (n - 1)d) + (a + (n - 2)d) +...….+ (a + d) + a\)

**……...…(2)**Now add

**(1)**and**(2)**then we get:\(2S_n =\) \([a + a + (n - 1)d] + [a + d + a + (n - 2)d] +.....+[a + (n - 2)d + (a + d)] + [a + ( n - 1)d + a]\)

\(2S_n =\) \([2a + (n - 1)d] + [2a + (n - 1)d] +....[2a + ( n - 1)d]\) [\(n\) times]

\(2S_n =\) \( n × [2a + (n - 1)d]\)

That is, ${S}_{n}=\frac{n}{2}(2a+(n-1)d)$

Therefore, the sum of first '\(n\)' terms of a Arithmetic Progression ${S}_{n}=\frac{n}{2}(2a+(n-1)d)$.

If the first term \(a\), and the last term '\(l\)' (\(n^t\)\(^h\) term) are given then, we can simplify the expression as follows.

$S=\frac{n}{2}[2a+(n-1)d]=\frac{n}{2}[a+a+(n-1)d]$

We know that \(l = a + ( n - 1)d\)

Therefore, the equation becomes $S=\frac{n}{2}[a+l]$. We can use this formula to find the sum \(S_n\), when the first term '\(a\)' and the last term '\(l\)' of the \(A.P\) series are the known values.

Important!

Suppose we have the series \(a_1, a_2, a_3,....a_n\)..... then the number of terms in the series is neither negative nor zero. That is the number of terms in the series is always positive.

Let's see an example to understand this concept much better.

Example:

**1**.

**Find the sum of the series**\(7, 13, 19, 25, 31,..\)

**upto**\(22\)

**terms**.

**Solution**:

The given series is \(7, 13, 19, 25, 31,..\)

Here, \(a = 7\), common difference \(d = 13 - 7 = 6\). And the total number of term \(n = 22\)

We know the formula to find the sum of the series when \(a, d\) and \(n\) values are known.

${S}_{n}=\frac{n}{2}(2a+(n-1)d)$

Let's substitute the known values in the formula.

$\begin{array}{l}{S}_{22}=\frac{22}{2}(2\times 7+(22-1)6)\\ \\ {S}_{22}=11(14+(21\times 6))\\ \\ {S}_{22}=11(14+126)\\ \\ {S}_{22}=11\times 140\\ \\ {S}_{22}=1540\end{array}$

Now we find the sum of the series using the formula $S=\frac{n}{2}[a+l]$, if the last value of the series given as \(133\).

Then the series will be series \(7, 13, 19, 25, 31,..133\).

Let's substitute '\(a\)' and '\(l\)' values in the formula.

$\begin{array}{l}{S}_{22}=\frac{22}{2}[7+133]\\ \\ {S}_{22}=11[140]\\ \\ {S}_{22}=1540\end{array}$

**Therefore**,

**the sum of given series is**\(1540\).