Find the sum of \[34 + 32 + 30 + ... + 10\].
Answer
549.9k+ views
Hint: In this question, we have to find out the sum from the given particular series.
We need to first find out the value of the first term and common difference. By subtracting the first term from the second term, we will get a common difference. Then put all the values in the formula of the $n^{th}$ term of the sequence to find the $n^{th}$ term of the sequence and then applying these to find out the sum of $n^{th}$ partial sum of the arithmetic sequence, we can find out the required solution.
Formula:
Property of A.P.:
The $n^{th}$ term of the Arithmetic sequence is
\[{a_n} = a + \left( {n - 1} \right)d\]
The $n^{th}$ partial sum of the Arithmetic sequence is
\[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\]
Where,
a = first term of the sequence
d = common difference
n= number of terms
Complete step by step answer:
We need to find out the value of the sum \[34 + 32 + 30 + ... + 10\].
a =first term of the sequence=\[34\].
d=the common difference=second term – first term=\[34 - 32 = - 2\].
Now we need to find out the value of n.
Since the difference between two consecutive numbers is the same, it is an arithmetic sequence.
We know that the $n^{th}$ term of the arithmetic sequence is
\[{a_n} = a + \left( {n - 1} \right)d\]
\[10 = 34 + (n - 1) \times - 2\]
\[2(n - 1) = 34 - 10\]
\[(n - 1) = \dfrac{{24}}{2} = 12\]
\[n = 12 + 1 = 13\]
n = number of terms =\[13\].
The sum of \[13\] th partial sum of the arithmetic sequence is
\[{S_{13}} = \dfrac{{13}}{2}\left[ {2 \times 34 + \left( {13 - 1} \right) \times \left( { - 2} \right)} \right]\]
Or,\[{S_{13}} = \dfrac{{13}}{2}\left[ {68 - 12 \times 2} \right]\]
Or, \[{S_{13}} = \dfrac{{13}}{2} \times \left[ {68 - 24} \right]\]
Or,\[{S_{13}} = 13 \times \dfrac{{44}}{2}\]
Or,\[{S_{13}} = 13 \times 22 = 286\]
Hence, \[34 + 32 + 30 + ... + 10\]=\[286\].
Note: An arithmetic progression is a sequence of numbers for which if we take the difference between any two successive numbers, we will get a constant.
In general, we write an Arithmetic sequence like this: \[\left\{ {a,a + d,a + 2d,a + 3d....} \right\}\] where a is the first term and d is the difference between the terms, called the common difference.
We need to first find out the value of the first term and common difference. By subtracting the first term from the second term, we will get a common difference. Then put all the values in the formula of the $n^{th}$ term of the sequence to find the $n^{th}$ term of the sequence and then applying these to find out the sum of $n^{th}$ partial sum of the arithmetic sequence, we can find out the required solution.
Formula:
Property of A.P.:
The $n^{th}$ term of the Arithmetic sequence is
\[{a_n} = a + \left( {n - 1} \right)d\]
The $n^{th}$ partial sum of the Arithmetic sequence is
\[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\]
Where,
a = first term of the sequence
d = common difference
n= number of terms
Complete step by step answer:
We need to find out the value of the sum \[34 + 32 + 30 + ... + 10\].
a =first term of the sequence=\[34\].
d=the common difference=second term – first term=\[34 - 32 = - 2\].
Now we need to find out the value of n.
Since the difference between two consecutive numbers is the same, it is an arithmetic sequence.
We know that the $n^{th}$ term of the arithmetic sequence is
\[{a_n} = a + \left( {n - 1} \right)d\]
\[10 = 34 + (n - 1) \times - 2\]
\[2(n - 1) = 34 - 10\]
\[(n - 1) = \dfrac{{24}}{2} = 12\]
\[n = 12 + 1 = 13\]
n = number of terms =\[13\].
The sum of \[13\] th partial sum of the arithmetic sequence is
\[{S_{13}} = \dfrac{{13}}{2}\left[ {2 \times 34 + \left( {13 - 1} \right) \times \left( { - 2} \right)} \right]\]
Or,\[{S_{13}} = \dfrac{{13}}{2}\left[ {68 - 12 \times 2} \right]\]
Or, \[{S_{13}} = \dfrac{{13}}{2} \times \left[ {68 - 24} \right]\]
Or,\[{S_{13}} = 13 \times \dfrac{{44}}{2}\]
Or,\[{S_{13}} = 13 \times 22 = 286\]
Hence, \[34 + 32 + 30 + ... + 10\]=\[286\].
Note: An arithmetic progression is a sequence of numbers for which if we take the difference between any two successive numbers, we will get a constant.
In general, we write an Arithmetic sequence like this: \[\left\{ {a,a + d,a + 2d,a + 3d....} \right\}\] where a is the first term and d is the difference between the terms, called the common difference.
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