Question

Which term of the AP 9, 14,19,24,29... is 379?

Answer 1

Hint: Find the differences between consecutive terms. You will see the sequence is an AP. Put the first term, common difference in the expression for ${{n}^{\text{th}}}$ term of an AP to obtain an equation in $n$. Solve it.

Complete step by step answer:
 A sequence is defined as the enumerated collection of numbers where repetitions are allowed and order of the numbers matters. It can also be expressed as a one-one map from the natural numbers set to real numbers. The members of the sequence are called terms. Mathematically, a sequence with infinite terms is written as
\[\left( {{x}_{n}} \right)={{x}_{1}},{{x}_{2}},{{x}_{3}},...\]
If the sequence has finite terms terminated by a term then we write the sequence as
\[\left( {{x}_{n}} \right)={{x}_{1}},{{x}_{2}},{{x}_{3}},...{{x}_{n}}\]
Arithmetic sequence otherwise known as arithmetic progression abbreviated as AP is a type sequence where the difference between any two consecutive numbers is constant . If $\left( {{x}_{n}} \right)$ is an AP, then ${{x}_{2}}-{{x}_{1}}={{x}_{3}}-{{x}_{2}}...$ . The difference between two terms is called common difference and denoted $d$ where $d={{x}_{2}}-{{x}_{1}}={{x}_{3}}-{{x}_{2}}...$. \[\]
Now
\[\begin{align}
  & d={{x}_{2}}-{{x}_{1}}\Rightarrow {{x}_{2}}={{x}_{1}}+d \\
 & d={{x}_{3}}-{{x}_{2}}\Rightarrow {{x}_{3}}={{x}_{2}}+d={{x}_{1}}+d+d={{x}_{1}}+2d \\
 & \vdots \\
\end{align}\]
 We observe the pattern and find out the expression for for ${{n}^{\text{th}}}$ term as
\[d={{x}_{1}}+\left( n-1 \right)d\]
Where ${{x}_{1}}$ is called the first term.
As given in the question the AP is with infinite terms 9,14,19,24,29,.... We can see here that the first term is 9, the second time term is 14 and the third term is 19 and so on. So ${{x}_{1}}=9$. The common difference of the AP is

\[d=14-9=19-14=24-19=29-24=5\]
We are asked to find out 379 which term of the given AP. So let assume that for ${{n}^{\text{th}}}$ term is the asked term and put these values in the expression for for ${{n}^{\text{th}}}$ term of an AP.
\[\begin{align}
  & {{x}_{n}}={{x}_{1}}+\left( n-1 \right)d \\
 & \Rightarrow 379=9+\left( n-1 \right)5 \\
 & \Rightarrow n-1=\dfrac{370}{5} \\
 & \Rightarrow n=74+1=75 \\
\end{align}\]

So 379 is the ${{75}^{\text{th}}}$ term of the given AP.\[\]

Note: The question can also be reframed to find the ${{n}^{\text{th}}}$ term of a geometric progression where there is a constant ratio between consecutive terms instead of constant difference. An AP series is the sum of the terms in AP sequence. The standard deviation of AP sequence is given by $\left| d \right|\sqrt{\dfrac{\left( n-1 \right)\left( n+1 \right)}{12}}.$