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(D) Given that ${a_1}, {a_2}, \dots, {a_{24}}$ are in an arithmetic progression $(A.P.)$.We know that in an $A.P.$,the sum of terms equidistant from t

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$900$

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(D) Given that ${a_1}, {a_2}, \dots, {a_{24}}$ are in an arithmetic progression $(A.P.)$.We know that in an $A.P.$,the sum of terms equidistant from the beginning and the end is constant and equal to the sum of the first and last terms,i.e.,${a_1} + {a_{24}} = {a_5} + {a_{20}} = {a_{10}} + {a_{15}}$.Given: ${a_1} + {a_5} + {a_{10}} + {a_{15}} + {a_{20}} + {a_{24}} = 225$.Substituting the property: $3({a_1} + {a_{24}}) = 225$.Therefore,${a_1} + {a_{24}} = \frac{225}{3} = 75$.The sum of the first $24$ terms is given by $S_{24} = \frac{n}{2}({a_1} + {a_n}) = \frac{24}{2}({a_1} + {a_{24}})$.$S_{24} = 12 	imes 75 = 900$.

If ${a_1}, {a_2}, {a_3}, \dots, {a_{24}}$ 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} + \dots + {a_{23}} + {a_{24}} = $

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