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

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(d) ${a_1} + {a_5} + {a_{10}} + {a_{15}} + {a_{20}} + {a_{24}} = 225$

$ \Rightarrow $ $({a_1} + {a_{24}}) + ({a_5} + {a_{20}}) + ({a_{10}} + {a_{15

Vedclass · Accepted Answer

$900$

Vedclass · Answer

(d) ${a_1} + {a_5} + {a_{10}} + {a_{15}} + {a_{20}} + {a_{24}} = 225$

$ \Rightarrow $ $({a_1} + {a_{24}}) + ({a_5} + {a_{20}}) + ({a_{10}} + {a_{15}}) = 225$

$ \Rightarrow $ $3({a_1} + {a_{24}}) = 225$

$ \Rightarrow $${a_1} + {a_{24}} = 75$

( In an $A.P.$ the sum of the terms equidistant from the beginning and the end is same and is equal to the sum of first and last term)

${a_1} + {a_2} + ...... + {a_{24}} = \frac{{24}}{2}({a_1} + {a_{24}}) = 12 	imes 75 = 900$.

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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