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

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Question

${a_1} + {a_2} + {a_3} = 39$

$ \Rightarrow {a_1} + \left( {{a_1} + d} \right) + \left( {{a_1} + 2d} \right) = 39$

$ \Rightarrow 3{a_1} + 3d = 39

Vedclass · Accepted Answer

$29.5$

Vedclass · Answer

${a_1} + {a_2} + {a_3} = 39$

$ \Rightarrow {a_1} + \left( {{a_1} + d} \right) + \left( {{a_1} + 2d} \right) = 39$

$ \Rightarrow 3{a_1} + 3d = 39\,\,\,\,\,\,\,\,\left[ {{a_1} = 10} \right]$

$ \Rightarrow d = 3$

Sum of last four term $=178$

Their mean $ = \frac{{178}}{4} = 44.5$

${a_n} = 44.5 + 1.5 + 3 = 49$

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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