Question

Find a49a49 of the sequence 70,63,56,49,…70,63,56,49,…. A. -243 B. -273 C. -63 D. -266

Answer 1

Well, from the sequence, notice 70,63,56,49,…

is simply dropping on the next term by 7 units, so,
70 - 7, 63
63 - 7, 56
and so on

thus, the "common difference", or the number you "add" to get the next term is -7..... as from the sequence itself, you can see the first term's value is 70.

$\bf n^{th}\textit{ term of an arithmetic sequence}\\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference}\\ ----------\\ a_1=70\\ d=-7\\ n=49 \end{cases} \\\\\\ a_{49}=70+(49-1)(-7)\implies a_{49}=70-336\implies a_{49}=-266$