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

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

Vedclass · Accepted Answer

$42$

Vedclass · Answer

(B) In an $A.P.$,the sum of terms equidistant from the beginning and end is constant.Specifically,$a_3 + a_{19} = a_5 + a_{17} = a_1 + a_{21} = 2a_{11}$.Let $a_3 + a_{19} = a_5 + a_{17} = 2a_{11} = k$.Given the equation: $a_3 + a_5 + a_{11} + a_{17} + a_{19} = 10$.Substituting the terms in terms of $k$: $\frac{k}{2} + \frac{k}{2} + \frac{k}{2} + \frac{k}{2} + \frac{k}{2} = 10$ is incorrect; rather,we have $(a_3 + a_{19}) + (a_5 + a_{17}) + a_{11} = 10$.This simplifies to $k + k + \frac{k}{2} = 10$,which gives $\frac{5k}{2} = 10$,so $k = 4$.Since $a_1 + a_{21} = 2a_{11} = k = 4$,the sum of the $21$ terms is given by $S_{21} = \frac{21}{2}(a_1 + a_{21})$.$S_{21} = \frac{21}{2} 	imes 4 = 42$.

If $a_1, a_2, a_3, .... a_{21}$ 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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