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8. Sequences and Series
hard
ધારો કે $a _1, a _2, \ldots, a _{2024}$ એક એવી સમાંતરશ્રેણી છે કે જેથી $a _1+\left( a _5+ a _{10}+ a _{15}+\ldots+ a _{2020}\right)+ a _{2024}= 2233$. તો $a_1+a_2+a_3+\ldots+a_{2024}$ ________
A$11157$
B$1574$
C$1156$
D$11132$
(JEE MAIN-2025)
Solution
$a_1+a_5+a_{10}+\ldots \ldots+a_{2020}+a_{2024}=2233$
In an $A.P.$ the sum of terms equidistant from end is equal.
$a_1+a_{2024}=a_5+a_{2020}=a_{10}+a_{2015} \cdots \cdots$
$\Rightarrow 203 \text { pairs }$
$\Rightarrow 203\left(a_1+a_{2024}\right)=2233$
Hence,
$S_{2024}=\frac{2024}{2}\left(a_1+a_{2024}\right)$
$= 1012 \times 11$
$= 11132$
In an $A.P.$ the sum of terms equidistant from end is equal.
$a_1+a_{2024}=a_5+a_{2020}=a_{10}+a_{2015} \cdots \cdots$
$\Rightarrow 203 \text { pairs }$
$\Rightarrow 203\left(a_1+a_{2024}\right)=2233$
Hence,
$S_{2024}=\frac{2024}{2}\left(a_1+a_{2024}\right)$
$= 1012 \times 11$
$= 11132$
Standard 11
Mathematics
