Let a_1, a_2, a_3, ..., a_n be in an A.P. If | vedclass

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(A) In an $A.P.$,the sum of terms equidistant from the beginning and the end is constant. Specifically,$a_k + a_{n-k+1} = a_1 + a_n$.Given the equatio

Vedclass · Accepted Answer

$306$

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(A) In an $A.P.$,the sum of terms equidistant from the beginning and the end is constant. Specifically,$a_k + a_{n-k+1} = a_1 + a_n$.Given the equation: $a_3 + a_7 + a_{11} + a_{15} = 72$.We observe that the indices sum to $3+15 = 18$ and $7+11 = 18$.Using the property $a_m + a_n = a_p + a_q$ if $m+n = p+q$,we have:$(a_3 + a_{15}) = (a_1 + a_{17})$ and $(a_7 + a_{11}) = (a_1 + a_{17})$.Substituting these into the given equation:$(a_1 + a_{17}) + (a_1 + a_{17}) = 72$$2(a_1 + a_{17}) = 72$$a_1 + a_{17} = 36$.The sum of the first $17$ terms of an $A.P.$ is given by $S_{17} = \frac{17}{2}(a_1 + a_{17})$.Substituting the value $a_1 + a_{17} = 36$:$S_{17} = \frac{17}{2} 	imes 36 = 17 	imes 18 = 306$.

Let $a_1, a_2, a_3, ..., a_n$ be in an $A.P.$ If $a_3 + a_7 + a_{11} + a_{15} = 72$,then the sum of its first $17$ terms is equal to:

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