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(C) Given the first term $a_1 = 10$ and the sum of the first three terms is $39.$$a_1 + (a_1 + d) + (a_1 + 2d) = 39$$3a_1 + 3d = 39$$3(10) + 3d = 39 \

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$29.5$

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(C) Given the first term $a_1 = 10$ and the sum of the first three terms is $39.$$a_1 + (a_1 + d) + (a_1 + 2d) = 39$$3a_1 + 3d = 39$$3(10) + 3d = 39 \Rightarrow 30 + 3d = 39 \Rightarrow 3d = 9 \Rightarrow d = 3.$Let the total number of terms be $n.$ The last four terms are $a_{n-3}, a_{n-2}, a_{n-1}, a_n.$Their sum is $(a_n - 3d) + (a_n - 2d) + (a_n - d) + a_n = 178.$$4a_n - 6d = 178.$Substituting $d = 3$: $4a_n - 6(3) = 178 \Rightarrow 4a_n - 18 = 178 \Rightarrow 4a_n = 196 \Rightarrow a_n = 49.$Since $a_n = a_1 + (n-1)d$,we have $49 = 10 + (n-1)3 \Rightarrow 39 = (n-1)3 \Rightarrow n-1 = 13 \Rightarrow n = 14.$For an $A.P.$ with $n = 14$ terms,the median is the average of the $\frac{n}{2}$-th and $(\frac{n}{2} + 1)$-th terms,i.e.,the $7$-th and $8$-th terms.Median $= \frac{a_7 + a_8}{2} = \frac{(a_1 + 6d) + (a_1 + 7d)}{2} = \frac{2a_1 + 13d}{2} = \frac{2(10) + 13(3)}{2} = \frac{20 + 39}{2} = \frac{59}{2} = 29.5$.

Let the sum of the first three terms of an $A.P.$ be $39$ and the sum of its last four terms be $178.$ If the first term of this $A.P.$ is $10,$ then the median of the $A.P.$ is

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