Given that $a_{4} + a_{8} + a_{12} + a_{16} = 224$ , the sum to nineteen terms of an A.P $a_{1}, a_{2}, a_{3} . . .$ is equal to ...........................

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Solution

Let $a_{10}$ be equal to a and the common difference be d.
Thus, we have $4 a = 224$ implying $a = 56.$
Now the nineteen terms will have the middle term as $a_{10}$ , which is $56$ .
Also, $a_{1} + a_{19} = 2 a_{10} = 112$
Such 9 pairs imply a sum of $112 \times 9$ i.e. $1008$
Adding the middle term $a_{10}$ gives the total required sum to be $1064$

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