Find the sum of the given series $45 + 46 + 47 + . . . + 113 + 114 + 115$

Open in App

Solution

Step 1: Given
The series is an arithmetic progression.
We can observe that the first term, $a = 45$ , $d = a_{2} - a_{1} = 46 - 45 = 1$ and $a_{n} = 115$
Step 2: Finding number of terms, $n$
The general form of an AP is,
$a_{n} = a + (n - 1) d$
Thus,
$115 = 45 + (n - 1) \cdot 1 \Rightarrow n - 1 = 70 \Rightarrow n = 71$
Thus, number of terms in the series is $71$
Step 3: Finding the sum of the series
The sum of an arithmetic progression is given as,
$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
Thus,
$S_{71} = \frac{71}{2} [2 \times 45 + (n - 1) \cdot 1] = 5680$
Therefore, the sum of the series $45 + 46 + 47 + . . . + 113 + 114 + 115$ is $5680$

Suggest Corrections

Similar questions

2 \sin^{- 1} \frac{3}{5} = \tan^{- 1} \frac{24}{7}

\tan^{- 1} \frac{1}{4} + \tan^{- 1} \frac{2}{9} = \frac{1}{2} \cos^{- 1} \frac{3}{5} = \frac{1}{2} \sin^{- 1} (\frac{4}{5})

\tan^{- 1} \frac{2}{3} = \frac{1}{2} \tan^{- 1} \frac{12}{5}

\tan^{- 1} \frac{1}{7} + 2 \tan^{- 1} \frac{1}{3} = \frac{π}{4}

\sin^{- 1} \frac{4}{5} + 2 \tan^{- 1} \frac{1}{3} = \frac{π}{2}

2 \sin^{- 1} \frac{3}{5} - \tan^{- 1} \frac{17}{31} = \frac{π}{4}

2 \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{8} = \tan^{- 1} \frac{4}{7}

2 \tan^{- 1} \frac{3}{4} - \tan^{- 1} \frac{17}{31} = \frac{π}{4}

2 \tan^{- 1} (\frac{1}{2}) + \tan^{- 1} (\frac{1}{7}) = \tan^{- 1} (\frac{31}{17})

4 \tan^{- 1} \frac{1}{5} - \tan^{- 1} \frac{1}{239} = \frac{π}{4}

| A | = ∣ ∣ ∣ ∣ \begin{matrix} 103 & 115 & 114 111 & 108 & 106 104 & 113 & 116 \end{matrix} ∣ ∣ ∣ ∣, | B | = ∣ ∣ ∣ ∣ \begin{matrix} 113 & 116 & 104 108 & 106 & 111 115 & 114 & 103 \end{matrix} ∣ ∣ ∣ ∣

1 + \frac{2}{11} + \frac{5}{11^{2}} + \frac{10}{11^{3}} + \frac{17}{11^{4}} + . .

∣ ∣ ∣ ∣ \begin{matrix} 103 & 115 & 114 111 & 108 & 106 104 & 113 & 116 \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} 113 & 116 & 104 108 & 106 & 111 115 & 114 & 103 \end{matrix} ∣ ∣ ∣ ∣

Join BYJU'S Learning Program