Let the sum of the first three terms of an A. | vedclass

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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$$3(10) + 3d = 39$ $\Rightarrow 30 + 3d = 39$ $\Rightarrow 3d = 9$ $\Rightarrow d = 3.$Let the number of terms be $n.$ The last four terms are $a_{n-3}, a_{n-2}, a_{n-1}, a_n.$Their sum is $4a_1 + ( (n-4) + (n-3) + (n-2) + (n-1) )d = 178.$$4(10) + (4n - 10)3 = 178$$40 + 12n - 30 = 178$ $\Rightarrow 12n + 10 = 178$ $\Rightarrow 12n = 168$ $\Rightarrow n = 14.$The median of an $A.P.$ with $n$ terms is the average of the first and last terms: $\frac{a_1 + a_n}{2}.$$a_n = a_1 + (n-1)d = 10 + (14-1)3 = 10 + 39 = 49.$Median $= \frac{10 + 49}{2} = \frac{59}{2} = 29.5.$

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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