Question

Consider the arithmetic sequence $6,10,14,.....$
(a) What is the sum of first $n$ consequence terms of the above sequence?
(b) How many consecutive terms from the beginning should be added to get the sum $240$?
(c) Is the sum of first-few consecutive terms becomes $250$? Why?

Answer 1

(a) $6,10,14,....$
Since $t_2-t_1=t_3-t_2$
Common difference, $d=10-6=4$
First term $a=6$
$S_n=\frac{n}{2}2a+(n-1)d=\frac{n}{2}2\times 6+(n-1)4=\frac{n}{2}12+4n-4=\frac{n}{2}4n+8$
$=n(2n+4)$
$2n^2+4n$
(b) For the sum to be $240$ from the beginning
$240=n^2+4n$
$\Rightarrow 240=n(2n+4)$
$\Rightarrow 240=n(2n+4)$
$\Rightarrow 120=n(n+2)$
$\Rightarrow n^2+2n-120=0$
$\Rightarrow n^2+12n-10n-120=0$
$\Rightarrow n(n+12)-10(n+12)=0$
$\Rightarrow (n+12)(n-10)=0$
$\therefore$ $n=-12,10$
$10$ terms should be added to make the sum to be $240$
(c) To make the sum to be $240$, '$n$' should be natural number.
$250=n(2n+4)$
$\Rightarrow 125=n(n+2)$
$\Rightarrow n^2+2n-125=0$
$\Rightarrow n^2+n-125=0$
$n=\frac{-1\sqrt{1-4\times 125}}{2}$
Since $n$ is not a natural number, therefore the term $250$ cannot exist in the series.