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(B) In an $A.P.$, the sum of terms equidistant from the beginning and the end is constant.Given the sum $a_1 + a_4 + a_7 + a_{10} + a_{13} + a_{16} =

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$98$

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(B) In an $A.P.$, the sum of terms equidistant from the beginning and the end is constant.Given the sum $a_1 + a_4 + a_7 + a_{10} + a_{13} + a_{16} = 147$.There are $6$ terms in this sequence. We can pair them as $(a_1 + a_{16}) + (a_4 + a_{13}) + (a_7 + a_{10}) = 147$.Since $a_1 + a_{16} = a_4 + a_{13} = a_7 + a_{10} = \lambda$, we have $3\lambda = 147$, which gives $\lambda = 49$.Now, we need to find $S = a_1 + a_6 + a_{11} + a_{16}$.Pairing these terms, we get $S = (a_1 + a_{16}) + (a_6 + a_{11})$.Since the sum of equidistant terms is constant in an $A.P.$, $a_1 + a_{16} = a_6 + a_{11} = \lambda = 49$.Therefore, $S = 49 + 49 = 98$.

If $ < a_n > $ is an $A.P.$ and $a_1 + a_4 + a_7 + .......+ a_{16} = 147$, then $a_1 + a_6 + a_{11} + a_{16}$ is equal to

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