
In an AP, \[{3^{rd}}\;\] term is $16$ and \[\]\[{7^{th}}\;\] term exceeds \[{5^{th}}\] term by $12$. Find the AP?
Answer
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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 \[{T_n}\; = a + \left( {n - 1} \right)d\], where \[{T_n}\; = {n^{th}}\;term\], \[a = \text{First term}\] and \[d = \text{Common difference} = {T_n} - {T_{n - 1}}\]. We should take care that the coefficient of d is always \[1\] less than the term number.
Complete step-by-step solution:
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 will write the given numbers in terms of a and d
So,\[{3^{rd}} \text{term} = a + 2d\]
\[{7^{th}}\text{term}\]=\[a + 6d\];
And ,5th term is
\[{5^{th}}\text{term}\]=\[a + 4d\]
We have given 3rd term is equal to 16,
So,
\[a + 2d = 16\;\] (1)
Since,\[{7^{th}}\text{term}\] exceeds \[{5^{th}}\text{term}\] by \[12\],
We will solve for d ,
\[a + 6d - \left\{ {a + 4d} \right\} = 12\]
now solving above equation,
\[a + 6d - a - 4d = 12\]
\[\Rightarrow 2d = 12\]
\[\Rightarrow d = 6\]
Common difference is 6.
We will now calculate the value of a by putting the value of d in equation \[1\],
\[a + 2d = 16\;\]
\[\Rightarrow a + 2 \times 6 = 16\]
By further solving,
\[\Rightarrow a + 12 = 16\]
\[\Rightarrow a = 4\]
so,
first term=4 and common difference=6
we have general form of A.P as
\[a,{\text{ }}a{\text{ }} + {\text{ }}d,{\text{ }}a{\text{ }} + {\text{ }}2d,{\text{ }}a{\text{ }} + {\text{ }}3d\]
We will substitute the values to get our AP,
Our required A.P is \[4,10,16{\text{ }} \ldots \ldots .\].
Note: An arithmetic progression is a set of numbers with a common difference between any two subsequent numbers (A.P.). A.P.'s example is \[\;3,6,9,12,15,18,21,{\text{ }}...\]
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.
Complete step-by-step solution:
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 will write the given numbers in terms of a and d
So,\[{3^{rd}} \text{term} = a + 2d\]
\[{7^{th}}\text{term}\]=\[a + 6d\];
And ,5th term is
\[{5^{th}}\text{term}\]=\[a + 4d\]
We have given 3rd term is equal to 16,
So,
\[a + 2d = 16\;\] (1)
Since,\[{7^{th}}\text{term}\] exceeds \[{5^{th}}\text{term}\] by \[12\],
We will solve for d ,
\[a + 6d - \left\{ {a + 4d} \right\} = 12\]
now solving above equation,
\[a + 6d - a - 4d = 12\]
\[\Rightarrow 2d = 12\]
\[\Rightarrow d = 6\]
Common difference is 6.
We will now calculate the value of a by putting the value of d in equation \[1\],
\[a + 2d = 16\;\]
\[\Rightarrow a + 2 \times 6 = 16\]
By further solving,
\[\Rightarrow a + 12 = 16\]
\[\Rightarrow a = 4\]
so,
first term=4 and common difference=6
we have general form of A.P as
\[a,{\text{ }}a{\text{ }} + {\text{ }}d,{\text{ }}a{\text{ }} + {\text{ }}2d,{\text{ }}a{\text{ }} + {\text{ }}3d\]
We will substitute the values to get our AP,
Our required A.P is \[4,10,16{\text{ }} \ldots \ldots .\].
Note: An arithmetic progression is a set of numbers with a common difference between any two subsequent numbers (A.P.). A.P.'s example is \[\;3,6,9,12,15,18,21,{\text{ }}...\]
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.
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