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The sum of $3^{rd}$ and $15^{th}$ elements of an arithmetic progression is equal to the sum of $6^{th}$, $11^{th}$ and $13^{th}$ elements of the same progression. Then which elements of the series should necessarily be equal to zero?
Given: The sum of $3^{rd}$ and $15^{th}$ elements of an arithmetic progression is equal to the sum of $6^{th}$, $11^{th}$ and $13^{th}$ elements of the same progression.
To do: To find which elements of the series should necessarily be equal to zero.
Solution:
Let the first term of AP be $a$ and difference be $d$.
We know that $n^{th}$ term, $a_n=a+( n-1)d$, where $a$ & $d$ are the first term and common difference of an AP.
$\therefore$ Then $3^{rd}$ term will be $a_3=a+2d$
$15^{th}\ term\ a_{15}=a+14d$
$6^{th}\ term\ a_{6}=a+5d$
$11^{th}\ term\ a_{11}=a+10d$
$13^{th}\ term\ a_{13}=a+12d$
According to the question,
$a_3+a_{15}=a_6+a_{11}+a_{13}$
$\Rightarrow a+2d+a+14d=a+5d+a+10d+a+12d$
$\Rightarrow 2a+16d=3a+27d$
$\Rightarrow a+11d=0$
As known $a+11d$ is the $12^{th}$ term of arithmetic progression.
Updated on: 10-Oct-2022
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