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Find the sum of first $24$ terms of the $\text{A.P.}\;a_1,\;a_2,\;a_3,\;\dots$ if it is known that $a_1+a_5+a_{10}+a_{15}+a_{20}+a_{24}=225$
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We know that in an $\text{A.P.}$, the sum of the terms equidistant from the beginning and end is always the same and is equal to the sum of the first and last term.

Therefore, $d=b-a$

$\Rightarrow a_1+a_{24}=a_5+a{20}=a_{10}+a_{15}$

It is given that

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

$\Rightarrow (a_1+a_24)+(a_1+a_{24})+(a+a_{24})=225$

$\Rightarrow 3(a_1+a_{24})=225$

$\Rightarrow a_1+a_{24}=75$

We know that

$\text{S}_n=\dfrac{n}{2}[a+l],$

Where $a$ is the first term, and $l$ is the last term of an $\text{A.P.}$

$\text{S}_{24}=\dfrac{24}{2}[a_1+a_{24}]=12\times 75=900.$
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