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Question: Find the first four terms in an A.P. when \[a = 10\] and \[d = 3\]....

Find the first four terms in an A.P. when a=10a = 10 and d=3d = 3.

Explanation

Solution

Here we will use the basics of the arithmetic progression to find the first four terms of the series. Arithmetic progression (A.P.) is the sequence of numbers such that the common difference between the consecutive numbers remains constant. We will put the value of the first term and common difference in the nthterm{n^{th}}{\rm{term}} formula to get the first four terms of the A.P.

Formula Used:
We will use the formula of nthterm{n^{th}}{\rm{term}} of the A.P. series is nthterm=a+(n1)d{n^{th}}{\rm{term}} = a + \left( {n - 1} \right)d, where aa is the first term, nn is the number of terms and dd is the common difference.

Complete step-by-step answer:
It is given that the first term a=10a = 10 and common difference d=3d = 3of the A.P. series.
Now we will put the value of the nn as 1,2,3 and 4 in nthterm=a+(n1)d{n^{th}}{\rm{term}} = a + \left( {n - 1} \right)d to get the first four value of the A.P. series.
Firstly we will put the value of n=1n = 1 to get the first term of the series. Therefore, we get
First term =10+(11)3 = 10 + \left( {1 - 1} \right)3
Subtracting the terms inside the bracket, we get
\RightarrowFirst term =10+(0)3=10 = 10 + \left( 0 \right)3 = 10
Now we will put the value of n=2n = 2 to get the second term of the series. So, we get
Second term =10+(21)3 = 10 + \left( {2 - 1} \right)3
Subtracting the terms inside the bracket, we get
\Rightarrow Second term =10+3=13 = 10 + 3 = 13
Now we will put the value of n=3n = 3 to get the third term of the series. So, we get
Third term =10+(31)3 = 10 + \left( {3 - 1} \right)3
Subtracting the terms inside the bracket, we get
\Rightarrow Third term =10+(2)3 = 10 + \left( 2 \right)3
Simplifying the terms, we get
\Rightarrow Third term =10+6=16 = 10 + 6 = 16
Now we will put the value of n=4n = 4 to get the fourth term of the series. So, we get
Fourth term =10+(41)3 = 10 + \left( {4 - 1} \right)3
Subtracting the terms inside the bracket, we get
\RightarrowFourth term =10+(3)3 = 10 + \left( 3 \right)3
Simplifying the terms, we get
\RightarrowFourth term =10+9=19 = 10 + 9 = 19
Hence, the first fourth term of the A.P. series is 10, 13, 16, 19.

Note: We used the terms arithmetic progression and common difference in the solution.
An arithmetic progression is a series of numbers in which each successive number is the sum of the previous number and a fixed difference. The fixed difference is called the common difference.
We calculated the common difference by subtracting the first term from the second term. The common difference is the fixed difference between each successive term of an A.P.
Therefore, we get
Common difference == Second term - First term == Third term - Second term == Fourth term - Third term