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8. Sequences and Series
medium
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}} = $
A
$909$
B
$75$
C
$750$
D
$900$
Solution
(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)
${a_1} + {a_2} + …… + {a_{24}} = \frac{{24}}{2}({a_1} + {a_{24}}) = 12 \times 75 = 900$.
Standard 11
Mathematics
