Question

An A.P is 4,9,14, 19... and so on. find its \[{10^{th}}\] term?

Answer 1

Hint: The general form of an Arithmetic Progression is a, a+ d, a + 2d, a+ 3d and so on. The formula for nth term of an AP series is \[{T_n} = {\text{a}} + \left( {n - 1} \right)d\] , where \[{T_n}\; = {n^{th}}\;term\] , \[a = first{\text{ }}term\] and \[d = common\,difference = {T_n}\; - {\text{ }}{T_{n - 1}}\] . We should take care that the coefficient of d is always \[1\] less than the term number.

Complete step-by-step answer:
 An arithmetic progression is a set of numbers with a common difference between any two subsequent numbers is always constant.
We assume a is the first term of an AP,
a = first term
we also assume d is the common difference between two consecutive numbers
$d = {T_n}\; - {\text{ }}{T_{n - 1}}$
We have given the series so first term is 4 and common difference is \[d = {T_n}\; - {\text{ }}{T_{n - 1}}\]
So, a=4 and \[d = {T_n}\; - {\text{ }}{T_{n - 1}}\]
Calculating the value of d
d=9-4
d=5
we will find the \[{n^{th}}\] by using
 \[{T_n}\; = a + \left( {n - 1} \right)d\]
so, for calculating the \[{10^{th}}\] term,
we substitute a, n and d
 \[{T_{10}}\; = {\text{ }}a{\text{ }} + {\text{ }}\left( {10{\text{ }} - {\text{ }}1} \right){\text{ }}d\]
 \[{T_{10}}\; = {\text{ }}4{\text{ }} + {\text{ }}\left( 9 \right){\text{ }}d\]
By, further solving
 \[{T_{10}}\; = {\text{ }}4 + {\text{ }}9 \times 5\]
 \[{T_{10}}\; = {\text{ }}4 + 45\]
 \[{T_{10}}\; = {\text{ }}49\]
So, the correct answer is “49”.

Note: A geometric progression is a sequence where every term bears a constant ratio to its preceding term. Geometric progression is a special type of sequence. In order to get the next term in the geometric progression, we have to multiply with a fixed term known as the common ratio, every time, and if we want to find the preceding term in the sequence, we just have to divide the term with the same common ratio unlike Arithmetic progression which has a common difference.