If a_1, a_2, a_3, dots, a_{21} are in A.P. an | vedclass

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(B) In an $A.P.$,the sum of terms equidistant from the beginning and end is constant,i.e.,$a_1 + a_{21} = a_3 + a_{19} = a_5 + a_{17} = 2a_{11}$.Given

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$42$

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(B) In an $A.P.$,the sum of terms equidistant from the beginning and end is constant,i.e.,$a_1 + a_{21} = a_3 + a_{19} = a_5 + a_{17} = 2a_{11}$.Given $a_3 + a_5 + a_{11} + a_{17} + a_{19} = 10$.Substituting the relations,we get $(a_3 + a_{19}) + (a_5 + a_{17}) + a_{11} = 10$.Since $a_3 + a_{19} = 2a_{11}$ and $a_5 + a_{17} = 2a_{11}$,the equation becomes $2a_{11} + 2a_{11} + a_{11} = 10$.$5a_{11} = 10 \Rightarrow a_{11} = 2$.We know that $a_1 + a_{21} = 2a_{11} = 2(2) = 4$.The sum of the first $21$ terms is given by $S_{21} = \frac{21}{2}(a_1 + a_{21})$.$S_{21} = \frac{21}{2}(4) = 21 	imes 2 = 42$.

If $a_1, a_2, a_3, \dots, a_{21}$ are in $A.P.$ and $a_3 + a_5 + a_{11} + a_{17} + a_{19} = 10$,then the value of $\sum_{r=1}^{21} a_r$ is:

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