Question

The sum of the first 19 terms of an arithmetic progression is equal to the twice of the value of 10th term. The value of 10th term will be ?

Answer 1

We know that,

$S_n=\dfrac {n}{2}(2a+(n-1)d)$

What about if $n$ is odd?!

Let $n=2m+1,\ \ m\in\mathbb{W}$

So, $m=\dfrac {n-1}{2}\quad\implies\quad m+1=\dfrac {n+1}{2}$

Then,

$S_n=\dfrac {n}{2}(2a+(2m+1-1)d)\\\\\\S_n=\dfrac {n}{2}\cdot 2(a+md)\\\\\\S_n=n(a+md)\\\\\\S_n=n\cdot a_{m+1}\\\\\\\large\boxed {S_n=n\cdot a_{\frac {n+1}{2}}}$

So this is a standard equation. By this equation,

$S_{19}=19\cdot a_{\frac {19+1}{2}}\\\\\\2\cdot a_{10}=19\cdot a_{10}\\\\\\17\cdot a_{10}=0\\\\\\\large\boxed {a_{10}=\mathbf {0}}$