Question

Find \[\text{k}\] given that \[3\text{k+1,k   and  }-3\] are the consecutive terms of an arithmetic sequence.

Answer 1

Hint: Using the formulas from Arithmetic sequence or arithmetic progression we can find the value of k. The nth term of the sequence is given by Tn=a+(n-1)d, where n is the nth term and a is the first term of the sequence and d is a common difference.

Complete step-by-step answer:

Let us take \[3\text{k+1,k   and  }-3\] as the terms  \[a,b\text{  and  }c\]

Since, \[a,b\text{  and  }c\] are the three consecutive terms we can say that...

\[nth\,\,\text{term}-(n-1)th\,\,\text{term}=\text{common difference}\]

Therefore, the methods and calculation are;

\[  b-a=\text{ common difference} \]

\[  c-b=\text{  common difference} \] 

\[ \therefore b-a=c-b \]

\[ b+b=a+c \]

\[ 2b=a+c......(1) \] 

\[ b=\text{k} \]

\[ a=3\text{k+1} \]

\[ c=-3 \] 

Therefore, by putting the values of \[a,b\text{  and  }c\] as \[3\text{k+1,k   and  }-3\] again in equation (1)

\[ 2(\text{k})=3\text{k+1-3} \]

\[ \text{3k}-\text{2k=}-1+3 \] 

\[ \text{k=2} \]

Therefore, we get the value of \[\text{k}\] as 2.

To verify we put the value of \[\text{k}\]in the consecutive terms given to us.

\[ 3\text{k+1}=3(2)+1=7 \]

\[k=2 \] and \[-3. \]

Therefore, \[7,2,-3\] are in arithmetic progression.

Note: The common difference of consecutive terms in arithmetic progression is always equal. We use this concept to find the value of K in the given equation by equating the difference between first and second terms and that of second and third terms.