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

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

After inserting $n$, $A.M.'s$ between $2$ and $38$, the sum of the resulting progression is $200$. The value of $n$ is

If $1,\,\,{\log _9}({3^{1 - x}} + 2),\,\,{\log _3}({4.3^x} - 1)$ are in $A.P.$ then $x$ equals

What is the sum of all two digit numbers which give a remainder of $4$ when divided by $6$ ?

The value of $x$ satisfying ${\log _a}x + {\log _{\sqrt a }}x + {\log _{3\sqrt a }}x + .........{\log _{a\sqrt a }}x = \frac{{a + 1}}{2}$ will be

If the sum of the series $2 + 5 + 8 + 11............$ is $60100$, then the number of terms are