Answer
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Hint: Let the first term of the sequence be $a$ and the common difference be $d$. Find the fourth term, eighth term, twelfth term and sixteenth term of the A.P. in terms of $a$ and $d$. Equate their sum to 224. Then, apply the formula of \[{S_n}\] and simplify it.
Complete step-by-step answer:
Let the first term of the sequence be $a$ and the common difference be $d$
${n^{th}}$ term of the sequence is given by the formula, ${a_n} = a + \left( {n - 1} \right)d$
Then, the fourth term can be written as, ${a_4} = a + 3d$.
Similarly, ${a_8} = a + 7d$, ${a_{12}} = a + 11d$ and ${a_{16}} = a + 15d$
We are given that the sum of fourth term, eighth term, twelfth term and sixteenth term of the A.P. is equal to 224.
On substituting the values of these terms we get,
$a + 3d + a + 7d + a + 11d + a + 15d = 224$
$4a + 36 = 224$
$a + 9d = 56$
We want to calculate the sum of the first nineteen terms of an A.P.
We will substitute $n = 19$ in the formula, ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$
${S_{19}} = \dfrac{19}{2}\left( {2a + \left( {19 - 1} \right)d} \right) = {S_{19}} = \dfrac{19}{2}\left( {2a + 18d} \right)$
On simplifying we get,
${S_{19}} = 19\left( {a + 9d} \right)$
We have already calculated the value of $a + 9d$ as 56.
Substitute it in ${S_{19}} =1 9\left( {a + 9d} \right)$ to find the required sum.
$
{S_{19}} =19\left( {56} \right) \\
{S_{19}} = 1064 \\
$
Hence, the sum of the first nineteen terms of an A.P. ${a_1},{a_2},{a_3}......$ if it is known that ${a_4} + {a_8} + {a_{12}} + {a_{16}} = 224$ is 1064.
Hence, option A is correct.
Note: The sum of $n$ terms of an A.P. is ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$, where $a$ is the first term, $d$ is the common difference and $n$ is the total number of terms. Also, the sum of $n$ terms of an A.P. is \[{S_n} = \dfrac{n}{2}\left( {a + {a_n}} \right)\], where \[{a_n}\] is the last term of the sequence.
Complete step-by-step answer:
Let the first term of the sequence be $a$ and the common difference be $d$
${n^{th}}$ term of the sequence is given by the formula, ${a_n} = a + \left( {n - 1} \right)d$
Then, the fourth term can be written as, ${a_4} = a + 3d$.
Similarly, ${a_8} = a + 7d$, ${a_{12}} = a + 11d$ and ${a_{16}} = a + 15d$
We are given that the sum of fourth term, eighth term, twelfth term and sixteenth term of the A.P. is equal to 224.
On substituting the values of these terms we get,
$a + 3d + a + 7d + a + 11d + a + 15d = 224$
$4a + 36 = 224$
$a + 9d = 56$
We want to calculate the sum of the first nineteen terms of an A.P.
We will substitute $n = 19$ in the formula, ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$
${S_{19}} = \dfrac{19}{2}\left( {2a + \left( {19 - 1} \right)d} \right) = {S_{19}} = \dfrac{19}{2}\left( {2a + 18d} \right)$
On simplifying we get,
${S_{19}} = 19\left( {a + 9d} \right)$
We have already calculated the value of $a + 9d$ as 56.
Substitute it in ${S_{19}} =1 9\left( {a + 9d} \right)$ to find the required sum.
$
{S_{19}} =19\left( {56} \right) \\
{S_{19}} = 1064 \\
$
Hence, the sum of the first nineteen terms of an A.P. ${a_1},{a_2},{a_3}......$ if it is known that ${a_4} + {a_8} + {a_{12}} + {a_{16}} = 224$ is 1064.
Hence, option A is correct.
Note: The sum of $n$ terms of an A.P. is ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$, where $a$ is the first term, $d$ is the common difference and $n$ is the total number of terms. Also, the sum of $n$ terms of an A.P. is \[{S_n} = \dfrac{n}{2}\left( {a + {a_n}} \right)\], where \[{a_n}\] is the last term of the sequence.
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