Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Answer
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Hint: Make the 1st equation using the given information that 3rd is 16 , (here we have to use nth term formula of an AP) , Now make another equation where the difference of 7th term and 5th term is 12 ( again use nth term formula for 7th and 5th term), solve them to find the AP . The nth term formula for an AP is : a+(n-1)d.
Complete step-by-step answer:
In the question we have to determine the arithmetic whose third term is 16 and the seventh term exceeds the 5th term by 12.
At first we briefly understand what is an arithmetic progression.
In mathematics, an arithmetic progression (AP) or an arithmetic sequence of numbers such that the difference between the consecutive terms is constant. Here difference means the second minus first. For instance, the sequence $5,7,9,11,13,15.....$ is an arithmetic progression with a common difference of 2.
If the initial term of an arithmetic is considered as a and common difference in successive term is d, then the \[{{n}^{th}}\] term of a sequence $\left( {{T}_{n}} \right)$ is denoted by :
${{T}_{n}}=a+\left( n-1 \right)d$
A finite portion of an arithmetic progression is called a finite arithmetic progression and sometimes called an arithmetic progression. The sum of finite arithmetic progression is called an arithmetic series.
In the question the 3rd term is given which is 16 so, we can write ${{T}_{3}}=16=a+\left( 3-1 \right)d$.
Where a is 1st term and d is a common difference.
So, $a+2d=16$
It is also said that the seventh term exceeds the fifth term by 12. So, ${{T}_{7}}-{{T}_{5}}=12$
$\left( a+6d \right)-\left( a+4d \right)=12$
On simplification we get,
$2d=12$
So, the common difference $\left( d \right)=6$.
We also know that $a+2d=16$ . So, $a+2\left( 6 \right)=16$.
Hence, on calculation we get 1st term $\left( a \right)$ as 4.
So, the arithmetic progression is $4,10,16,22....$
The A.P. is $4,10,16,22....$
Note: The behaviour of the arithmetic progression depends on common difference d. If the common difference is positive, then the terms will grow towards positive infinity and if the common difference is negative then the terms will grow towards negative infinity.
Complete step-by-step answer:
In the question we have to determine the arithmetic whose third term is 16 and the seventh term exceeds the 5th term by 12.
At first we briefly understand what is an arithmetic progression.
In mathematics, an arithmetic progression (AP) or an arithmetic sequence of numbers such that the difference between the consecutive terms is constant. Here difference means the second minus first. For instance, the sequence $5,7,9,11,13,15.....$ is an arithmetic progression with a common difference of 2.
If the initial term of an arithmetic is considered as a and common difference in successive term is d, then the \[{{n}^{th}}\] term of a sequence $\left( {{T}_{n}} \right)$ is denoted by :
${{T}_{n}}=a+\left( n-1 \right)d$
A finite portion of an arithmetic progression is called a finite arithmetic progression and sometimes called an arithmetic progression. The sum of finite arithmetic progression is called an arithmetic series.
In the question the 3rd term is given which is 16 so, we can write ${{T}_{3}}=16=a+\left( 3-1 \right)d$.
Where a is 1st term and d is a common difference.
So, $a+2d=16$
It is also said that the seventh term exceeds the fifth term by 12. So, ${{T}_{7}}-{{T}_{5}}=12$
$\left( a+6d \right)-\left( a+4d \right)=12$
On simplification we get,
$2d=12$
So, the common difference $\left( d \right)=6$.
We also know that $a+2d=16$ . So, $a+2\left( 6 \right)=16$.
Hence, on calculation we get 1st term $\left( a \right)$ as 4.
So, the arithmetic progression is $4,10,16,22....$
The A.P. is $4,10,16,22....$
Note: The behaviour of the arithmetic progression depends on common difference d. If the common difference is positive, then the terms will grow towards positive infinity and if the common difference is negative then the terms will grow towards negative infinity.
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