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Find the average of first 97 natural numbers.

147
237
348
449

Answer:- 1

Explanation:-

Solution:

1^st Method:
Average of 1^st n natural number is given by

= \frac{(\frac{[n \times (n + 1)]}{2})}{n}

$= \frac{{\left( {\frac{{\left[ {{\text{n}} \times \left( {{\text{n}} + 1} \right)} \right]}}{2}} \right)}}{{\text{n}}}$
Average of 1^st 97 natural number is given by

\begin{aligned} {\frac{(\frac{[97 \times (97 + 1)]}{2})}{97}} \\ = 49 \end{aligned}

$\eqalign{ & \left\{ {\frac{{\left( {\frac{{\left[ {97 \times \left( {97 + 1} \right)} \right]}}{2}} \right)}}{{97}}} \right\} \cr & = 49 \cr}$

2^st Method:
These numbers are in AP series so average.

\begin{aligned} = \frac{(sum of corresponding term)}{2} \\ = \frac{(1 + 97)}{2} = 49 \\ or, \frac{(2 + 96)}{2} = 49 \\ or, \frac{(3 + 95)}{2} = 49 \\ And so on . \end{aligned}

$\eqalign{ & = \frac{{\left( {{\text{sum of corresponding term}}} \right)}}{2} \cr & = \frac{{\left( {1 + 97} \right)}}{2} = 49 \cr & {\text{or, }}\frac{{\left( {2 + 96} \right)}}{2} = 49 \cr & {\text{or, }}\frac{{\left( {3 + 95} \right)}}{2} = 49 \cr & {\text{And so on}}{\text{.}} \cr}$

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Average of 1^st 97 natural number is given by

\begin{aligned} {\frac{(\frac{[97 \times (97 + 1)]}{2})}{97}} \\ = 49 \end{aligned}

$\eqalign{ & \left\{ {\frac{{\left( {\frac{{\left[ {97 \times \left( {97 + 1} \right)} \right]}}{2}} \right)}}{{97}}} \right\} \cr & = 49 \cr}$

2^st Method:
These numbers are in AP series so average.

\begin{aligned} = \frac{(sum of corresponding term)}{2} \\ = \frac{(1 + 97)}{2} = 49 \\ or, \frac{(2 + 96)}{2} = 49 \\ or, \frac{(3 + 95)}{2} = 49 \\ And so on . \end{aligned}

$\eqalign{ & = \frac{{\left( {{\text{sum of corresponding term}}} \right)}}{2} \cr & = \frac{{\left( {1 + 97} \right)}}{2} = 49 \cr & {\text{or, }}\frac{{\left( {2 + 96} \right)}}{2} = 49 \cr & {\text{or, }}\frac{{\left( {3 + 95} \right)}}{2} = 49 \cr & {\text{And so on}}{\text{.}} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

1^st Method:
Average of 1^st n natural number is given by

= \frac{(\frac{[n \times (n + 1)]}{2})}{n}

$= \frac{{\left( {\frac{{\left[ {{\text{n}} \times \left( {{\text{n}} + 1} \right)} \right]}}{2}} \right)}}{{\text{n}}}$
Average of 1^st 97 natural number is given by

\begin{aligned} {\frac{(\frac{[97 \times (97 + 1)]}{2})}{97}} \\ = 49 \end{aligned}

$\eqalign{ & \left\{ {\frac{{\left( {\frac{{\left[ {97 \times \left( {97 + 1} \right)} \right]}}{2}} \right)}}{{97}}} \right\} \cr & = 49 \cr}$

2^st Method:
These numbers are in AP series so average.

\begin{aligned} = \frac{(sum of corresponding term)}{2} \\ = \frac{(1 + 97)}{2} = 49 \\ or, \frac{(2 + 96)}{2} = 49 \\ or, \frac{(3 + 95)}{2} = 49 \\ And so on . \end{aligned}

$\eqalign{ & = \frac{{\left( {{\text{sum of corresponding term}}} \right)}}{2} \cr & = \frac{{\left( {1 + 97} \right)}}{2} = 49 \cr & {\text{or, }}\frac{{\left( {2 + 96} \right)}}{2} = 49 \cr & {\text{or, }}\frac{{\left( {3 + 95} \right)}}{2} = 49 \cr & {\text{And so on}}{\text{.}} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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