Question

find the value of `k in the arithmetic sequence. (3k-4), (7k+1), (12k -5)Write the sequence.​

Answer 1

Step-by-step explanation:

We have,

$\rm (3k-4), (7k+1), (12k-5),...$

$\rm d = (7k+1)-(3k-4)$

$\rm d = 7k+1-3k+4$

$\rm d = 4k+5$

d is also equals to,

$\rm d = (12k-5)-(7k+1)$

$\rm \implies 12k-7k-5-1$

$\rm \implies 5k-6$

so,

$\rm 5k-6=4k+5$

$\rm 5k-4k=5+6$

$\rm \implies k=11$

$\rm a_1= 3k-4 \Rightarrow 3(11)-4=29$

$\rm a_2 = 7k+1 \Rightarrow 7(11)+1=78$

$\rm a_3 = 12k-5 \Rightarrow 12(11)-5=127$

So, the sequence is,

$\rm 29,78,127,...$