
Which term of the AP: \[3,\text{ }8,\text{ }13,\text{ }18,\text{ }.\text{ }.\text{ }.\text{ },\text{ }is\text{ }78\]?
Answer
528.9k+ views
Hint: In order to find solution to this Arithmetic Progression Problem, we have to use a formula for finding the $n-th$ term of an Arithmetic Progression that is ${{a}_{n}}=a+\left( n-1 \right)d$ , where ${{a}_{n}}$ is the value of the ${{n}^{th}}$ term, $a$ is the initial term, $n$ is the total number of terms and $d$ is the common difference, to find which term in this Arithmetic Series is $78$.
Complete step by step solution:
We have our given series as \[3,\text{ }8,\text{ }13,\text{ }18,\text{ }.\text{ }.\text{ }.\text{ },78\].
With this we have to find which term is $78$.
Therefore, we will apply a formula for finding the ${{n}^{th}}$ term of an Arithmetic Progression that is ${{a}_{n}}=a+\left( n-1 \right)d$ .
With this, we get:
Last term in an Arithmetic series, ${{a}_{n}}=78$
First term, $a=3$
Common difference, $d={{a}_{2}}-{{a}_{1}}=8-3=5$
${{n}^{th}}$ term we have to find, $n=?$
Since we have all we got to evaluate into the formula, therefore, putting values into these formulas, we get:
${{a}_{n}}=a+\left( n-1 \right)d$
On evaluating, we get:
$\Rightarrow 78=3+\left( n-1 \right)\times 5$
On taking $+3$ on left-hand side, we get our expression as:
$\Rightarrow 78-3=\left( n-1 \right)\times 5$
On simplifying, we get:
$\Rightarrow 75=\left( n-1 \right)\times 5$
Now, on taking $5$ on left-hand side and applying sign rule, we get our expression as:
$\Rightarrow \dfrac{75}{5}=\left( n-1 \right)$
On simplifying, we get:
$\Rightarrow 15=\left( n-1 \right)$
On further simplification and eliminating brackets, we get our expression as:
$\Rightarrow 15=n-1$
Now, on taking $-1$ on left-hand side and applying sign rule, we get:
$\Rightarrow 15+1=n$
On Simplifying, we get:
$\Rightarrow n=16$
Therefore, the ${{16}^{th}}$ term of an Arithmetic Progression is $78$.
Note: Arithmetic Progression (AP) is a sequence of numbers in order in which the difference of any two consecutive numbers is a constant value.
We have two major formulas which is related to ${{n}^{th}}$ term of Arithmetic Progression:
To find the ${{n}^{th}}$ term of A.P: ${{a}_{n}}=a+\left( n-1 \right)d$
To find sum of ${{n}^{th}}$ term of A.P: $S=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$
Based on the given question of an Arithmetic Progression, we have to decide which formula we have to use.
Complete step by step solution:
We have our given series as \[3,\text{ }8,\text{ }13,\text{ }18,\text{ }.\text{ }.\text{ }.\text{ },78\].
With this we have to find which term is $78$.
Therefore, we will apply a formula for finding the ${{n}^{th}}$ term of an Arithmetic Progression that is ${{a}_{n}}=a+\left( n-1 \right)d$ .
With this, we get:
Last term in an Arithmetic series, ${{a}_{n}}=78$
First term, $a=3$
Common difference, $d={{a}_{2}}-{{a}_{1}}=8-3=5$
${{n}^{th}}$ term we have to find, $n=?$
Since we have all we got to evaluate into the formula, therefore, putting values into these formulas, we get:
${{a}_{n}}=a+\left( n-1 \right)d$
On evaluating, we get:
$\Rightarrow 78=3+\left( n-1 \right)\times 5$
On taking $+3$ on left-hand side, we get our expression as:
$\Rightarrow 78-3=\left( n-1 \right)\times 5$
On simplifying, we get:
$\Rightarrow 75=\left( n-1 \right)\times 5$
Now, on taking $5$ on left-hand side and applying sign rule, we get our expression as:
$\Rightarrow \dfrac{75}{5}=\left( n-1 \right)$
On simplifying, we get:
$\Rightarrow 15=\left( n-1 \right)$
On further simplification and eliminating brackets, we get our expression as:
$\Rightarrow 15=n-1$
Now, on taking $-1$ on left-hand side and applying sign rule, we get:
$\Rightarrow 15+1=n$
On Simplifying, we get:
$\Rightarrow n=16$
Therefore, the ${{16}^{th}}$ term of an Arithmetic Progression is $78$.
Note: Arithmetic Progression (AP) is a sequence of numbers in order in which the difference of any two consecutive numbers is a constant value.
We have two major formulas which is related to ${{n}^{th}}$ term of Arithmetic Progression:
To find the ${{n}^{th}}$ term of A.P: ${{a}_{n}}=a+\left( n-1 \right)d$
To find sum of ${{n}^{th}}$ term of A.P: $S=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$
Based on the given question of an Arithmetic Progression, we have to decide which formula we have to use.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

