Question

Which term of the AP: 31, 28, 25, ... is the first negative term.

Answer 1

Hint: Calculate the common difference of the given AP by subtracting any two consecutive terms. Assume that the first negative term is the ${{n}^{th}}$ term. Use the formula ${{a}_{n}}=a+\left( n-1 \right)d$ to write the ${{n}^{th}}$ term of the AP. Write an inequality for this term using the fact that it is negative. Simplify the inequality to calculate the value of ‘n’ substitute the value of ‘n’ in the formula to calculate the first negative term.

Complete step-by-step answer:
We have to calculate the first negative term of the given AP 31, 28, 25, ...
We will first calculate the common difference of the given AP. To do so, we will subtract any two consecutive terms. Thus, the common difference is $=28-31=-3$.
Let’s assume that the first negative term is the ${{n}^{th}}$ term.
We know the formula for writing the ${{n}^{th}}$ term is ${{a}_{n}}=a+\left( n-1 \right)d$, where ‘a’ is the first term, ‘n’ is the number of terms and ‘d’ is the common difference.
Substituting \[a=31,d=-3\] in the above expression, we have ${{a}_{n}}=31+\left( n-1 \right)\left( -3 \right)$.
Simplifying the above expression, we have ${{a}_{n}}=31-3n+3=34-3n$.
We know that ${{a}_{n}}$ is the first negative term. Thus, we have ${{a}_{n}}<0$.
So, we have $34-3n<0$.
Simplifying the above expression, we have $n>\dfrac{34}{3}$. The least positive integral value of ‘n’ which satisfies the given equation is $n=12$.
Substituting $n=12$ in the expression ${{a}_{n}}=31-3n+3=34-3n$, we have ${{a}_{12}}=34-3\left( 12 \right)=34-36=-2$.
Hence, the first negative term is the ${{12}^{th}}$ term, which is -2.

Note: Arithmetic Progression (AP) is a sequence of numbers in which the difference between any two consecutive terms is a constant. One must remember that the value of ‘n’ is always a positive integer. As the common difference of this AP is negative, we observe that it’s a decreasing AP.