Browse Questions

Find the value of $n$ such that the $n^{th}$ term of the sequences 27,24,21------- and 2,4,6----- are equal.

$\begin{array}{1 1} n=5 \\ n=6 \\ n=7 \\ n=4 \end{array}$

Explanation : $a_{1}=27\;,d_{1}=-3\;,t_{n}=27-3(n-1)$
$a_{2}=2\;,d_{2}=2\;,t_{n}=2+2(n-1)$
$Since \;n^{th}\;terms\;are\;equal$
$2+2(n-1)=27-3(n-1)$
$5(n-1)=25$
$n-1=5$
$n=6.$