If {a_1}, {a_2}, {a_3}, ..., {a_{24}} are in | vedclass

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(D) In an arithmetic progression $(A.P.)$,the sum of terms equidistant from the beginning and the end is constant and equal to the sum of the first an

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$900$

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(D) In an arithmetic progression $(A.P.)$,the sum of terms equidistant from the beginning and the end is constant and equal to the sum of the first and last terms.Given: ${a_1} + {a_5} + {a_{10}} + {a_{15}} + {a_{20}} + {a_{24}} = 225$.We know that in an $A.P.$,${a_1} + {a_{24}} = {a_5} + {a_{20}} = {a_{10}} + {a_{15}}$.Substituting these into the given equation:$3({a_1} + {a_{24}}) = 225$${a_1} + {a_{24}} = 75$.The sum of the first $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}(a_1 + a_n)$.For $n = 24$:$S_{24} = \frac{24}{2}({a_1} + {a_{24}}) = 12 	imes 75 = 900$.

If ${a_1}, {a_2}, {a_3}, ..., {a_{24}}$ are in arithmetic progression and ${a_1} + {a_5} + {a_{10}} + {a_{15}} + {a_{20}} + {a_{24}} = 225$,then ${a_1} + {a_2} + {a_3} + ... + {a_{23}} + {a_{24}} = $

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