
Calculate the sum of even numbers between 100 and 150 which are divisible by 13.
Answer
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Hint: In these types of questions which the given series is in arithmetic series we should be using sum of n terms in the arithmetic series formula that can be can be written as, ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$, and also we will make use of the formula for number of terms in an arithmetic series i.e., ${T_n} = a + \left( {n - 1} \right)d$.
Complete answer:
Given numbers are 100 and 150, we have to find the sum of numbers between 100 and 150 which are divisible by 13.
The numbers that are divisible by 13 between 100 and 150 will be $104,117,....143$,
The sum of these numbers can be calculated by using arithmetic series formula i.e.,
The sum of $n$ terms in an arithmetic series can be written as,
${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$,
Where$n$is the number of terms,$a$is the first term and $d$ is the common difference.
Here from the above given data,
First term, $a = 104$,
Common difference, $d = 13$,
Last term ${T_n} = 104$
And now we have to first find the number of terms i.e.,$n$ using last term formula i.e.,${T_n} = a + \left( {n - 1} \right)d$,
Now substituting the values we get,
${T_n} = a + \left( {n - 1} \right)d$,
$143 = 104 + \left( {n - 1} \right)13$,
$ \Rightarrow 143 = 104 + 13n - 13$,
Now adding the constant terms on R.H.S we get,
$ \Rightarrow 143 = 91 + 13n$,
Now taking the constant terms to one side we get,
$ \Rightarrow 13n = 143 - 91$
Now subtracting we get,
$ \Rightarrow 13n = 52$
Now multiplying and dividing both sides with 13 we get,
$\dfrac{{13n}}{{13}} = \dfrac{{52}}{{13}}$
Now by simplifying we get,
$n = 4$
Now substituting all the values in the sum of $n$ terms in an arithmetic series formula which is given as,
${S_n} = \dfrac{1}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$,
Here $a = 104$,$d = 13$,and $n = 4$, we get,
${S_n} = \dfrac{4}{2}\left[ {2\left( {104} \right) + \left( {4 - 1} \right)13} \right]$
Now simplifying we get,
${S_n} = 2\left[ {208 + \left( 3 \right)13} \right]$,
By multiplying the terms we get,
${S_n} = 2\left[ {208 + 39} \right]$,
Now again simplifying we get,
${S_n} = 2\left[ {247} \right]$,
Now multiplying we get,
${S_n} = 494$
$\therefore $The sum of even numbers between 100 and 150 which are divisible by 13 is 494.
The sum of even numbers between 100 and 150 which are divisible by 13 is 494.
Note:
As there are 3 types of series i.e., Arithmetic series, Geometric series and Harmonic series, students should not get confused which series to be used in what type of questions, as there are many formulas related to each of the series, here are some useful formulas related to the above series,
Sum of the $n$ terms in A. P is given by,${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$, where \[n\] is common ratio, \[a\] is the first term.
The \[{n^{th}}\] term In A.P is given by ${T_n} = a + \left( {n - 1} \right)d$,
Complete answer:
Given numbers are 100 and 150, we have to find the sum of numbers between 100 and 150 which are divisible by 13.
The numbers that are divisible by 13 between 100 and 150 will be $104,117,....143$,
The sum of these numbers can be calculated by using arithmetic series formula i.e.,
The sum of $n$ terms in an arithmetic series can be written as,
${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$,
Where$n$is the number of terms,$a$is the first term and $d$ is the common difference.
Here from the above given data,
First term, $a = 104$,
Common difference, $d = 13$,
Last term ${T_n} = 104$
And now we have to first find the number of terms i.e.,$n$ using last term formula i.e.,${T_n} = a + \left( {n - 1} \right)d$,
Now substituting the values we get,
${T_n} = a + \left( {n - 1} \right)d$,
$143 = 104 + \left( {n - 1} \right)13$,
$ \Rightarrow 143 = 104 + 13n - 13$,
Now adding the constant terms on R.H.S we get,
$ \Rightarrow 143 = 91 + 13n$,
Now taking the constant terms to one side we get,
$ \Rightarrow 13n = 143 - 91$
Now subtracting we get,
$ \Rightarrow 13n = 52$
Now multiplying and dividing both sides with 13 we get,
$\dfrac{{13n}}{{13}} = \dfrac{{52}}{{13}}$
Now by simplifying we get,
$n = 4$
Now substituting all the values in the sum of $n$ terms in an arithmetic series formula which is given as,
${S_n} = \dfrac{1}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$,
Here $a = 104$,$d = 13$,and $n = 4$, we get,
${S_n} = \dfrac{4}{2}\left[ {2\left( {104} \right) + \left( {4 - 1} \right)13} \right]$
Now simplifying we get,
${S_n} = 2\left[ {208 + \left( 3 \right)13} \right]$,
By multiplying the terms we get,
${S_n} = 2\left[ {208 + 39} \right]$,
Now again simplifying we get,
${S_n} = 2\left[ {247} \right]$,
Now multiplying we get,
${S_n} = 494$
$\therefore $The sum of even numbers between 100 and 150 which are divisible by 13 is 494.
The sum of even numbers between 100 and 150 which are divisible by 13 is 494.
Note:
As there are 3 types of series i.e., Arithmetic series, Geometric series and Harmonic series, students should not get confused which series to be used in what type of questions, as there are many formulas related to each of the series, here are some useful formulas related to the above series,
Sum of the $n$ terms in A. P is given by,${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$, where \[n\] is common ratio, \[a\] is the first term.
The \[{n^{th}}\] term In A.P is given by ${T_n} = a + \left( {n - 1} \right)d$,
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