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# Let ${x_1},{x_2},..........{x_n}$ be in an AP of ${x_1} + {x_4} + {x_9} + {x_{11}} + {x_{20}} + {x_{22}} + {x_{27}} + {x_{30}} = 272$ then ${x_1} + {x_2} + {x_3} + ........... + {x_{30}}$ is equal to A. 1020B. 1200C. 716D. 2720

Last updated date: 12th Aug 2024
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Hint: Here we need to solve the given question by considering the property of an AP i.e., ${a_1} + {a_l} = {a_2} + {a_{l - 2}} = {a_3} + {a_{l - 3}}$

Given that AP consists of 30 terms.
We know a property of an AP that ${a_1} + {a_l} = {a_2} + {a_{l - 2}} = {a_3} + {a_{l - 3}}$ here ${a_l}$ is the last term of an AP.
By above property we can write ${x_1} + {x_{30}} = {x_4} + {x_{27}} = {x_9} + {x_{22}} = {x_{11}} + {x_{20}}$
$\because {x_1} + {x_4} + {x_9} + {x_{11}} + {x_{20}} + {x_{22}} + {x_{27}} + {x_{30}} = 272 \\ \Rightarrow ({x_1} + {x_{30}}) + ({x_4} + {x_{27}}) + ({x_9} + {x_{22}}) + ({x_{11}} + {x_{20}}) = 272 \\ \Rightarrow 4({x_1} + {x_{30}}) = 272{\text{ }}\because {x_1} + {x_{30}} = {x_4} + {x_{27}} = {x_9} + {x_{22}} = {x_{11}} + {x_{20}} \\ \Rightarrow ({x_1} + {x_{30}}) = \dfrac{{272}}{4} = 68 \\$

Here we find the sum of the first term and last term of given AP. Now we can apply the formula of summation of n terms of AP.

${S_n} = \dfrac{n}{2}(a + l)$ Here n=number of terms, a = first term, l=last term
Now by substituting the values
${S_{30}} = \dfrac{{30}}{2}({x_1} + {x_{30}}) \\ {S_{30}} = \dfrac{{30}}{2}(68){\text{ }}\because {x_1} + {x_{30}} = 68 \\ \therefore {S_{30}} = 1020 \\$

Here the correct option is A.

Note: This is the question on the formula of summation of AP. We have to find the sum of the first and last term as required in formula so we apply the property of AP and find the sum of first and last term. By putting this in formula we get the answer.