Let be an Arithmetic Progression such that . Then is equal to __________. [2025]

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Solution

Q.
Let $a_{1}, a_{2}, . . ., a_{2024}$ be an Arithmetic Progression such that $a_{1} + (a_{5} + a_{10} + a_{15} + . . . + a_{2020}) + a_{2024} = 2233$ . Then $a_{1} + a_{2} + a_{3} + . . . + a_{2024}$ is equal to __________. [2025]

Ans.
(11132)

We have,

$a_{1} + a_{5} + a_{10} + a_{15} + . . . + a_{2020} + a_{2024} = 2233$ ... (i)

Since, $a_{1}, a_{2}, . . ., a_{2024}$ be in A.P.

$∴ a_{1} + a_{2024} = a_{5} + a_{2020} = a_{10} + a_{2015} . . . . .$

{Sum of terms equidistant form ends is equal}

From (i), we get

$(a_{1} + a_{2024}) + (a_{5} + a_{2020}) + (a_{10} + a_{2015}) + . . . = 2233$

$\Rightarrow 203 (a_{1} + a_{2024}) = 2233$ [ $∵$ Total pairs = 203]

$\Rightarrow a_{1} + a_{2024} = 11$

$∴$ Sum of n terms in A.P. is

$S_{n} = \frac{n}{2} (a_{1} + a_{2024})$

$= 1012 \times 11$ [ $∵$ n = 2024]

$∴$ $S_{n} = 11132$ .

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