Question

Third term of an arithmetic sequence is $34$ and its eighth term is $69.$
(a) Find the common difference of this sequence
(b) Write the algebraic form of this sequence
(c) If a new sequence is formed by multiplying each term of the given sequence by $4$ and the adding $3$ , what is the tenth term of the new sequence so formed?

Answer 1

$t_3=34$ and $t_8=69$
(a) Now, $t_n=a+(n-1)d$
$\Rightarrow t_3=a+(2-1)d$ and $t_8=a+(7-1)d$
$\Rightarrow 34=a+(2-1)d$ and $69=a+(7-1)d$
$\Rightarrow 34=a+d$ and $69=a+6d$
Thus, $69-34=a+6d-a-d$
$\Rightarrow 35=5d\Rightarrow d=7$
(b) First term, $a=t_3-2d=34-2\left(7\right)=34-14=20$
General term, $t_n=a+(n-1)d=20+(n-1)7=20+7n-7=7n+13$
Thus, $t_n=7n+13$ is the algebraic form of this sequence.
(c) $t_n=7n+13$
If each term of the sequence is multiplied by 4 and then 3 is added to it, the
general term will be $t_n=\left[\right(7n+13)\times 4]+3=28n+52+3=28n+55$
Thus, tenth term, $t_{10}=28\left(10\right)+55=280+55=335$