
If \[2x\], \[x + 8\], \[3x + 1\] are in A.P. then the value of \[x\] will be
A. 3
B. 7
C. 5
D. -2
Answer
512.1k+ views
Hint: Here in this question, given a sequence of arithmetic progression (A.P) we have to find the value of \[x\] in the terms of A.P. As we know the common difference between any two terms in Arithmetic progression is the same; it can be found by subtracting 1st term from 2nd term or 2nd term by 3rd term and hence equating both the expressions to get the required value of \[x\].
Complete step by step answer:
The general arithmetic progression is of the form \[a,a + d,a + 2d,...\] where $a$ is first term nth $d$ is the common difference which is the same between the distance on any two numbers in sequence otherwise the fixed number that must be added to any term of an AP to get the next term is known as the common difference of the AP.
The nth term of the arithmetic progression is defined as \[{a_n} = a + (n - 1)d\]. The common difference ‘\[d\]’ can be determined by subtracting the first term with the second term, second term with the third term, and so forth. Now let us consider the A.P. \[2x\], \[x + 8\], \[3x + 1\] we have to find the value of \[x\].
Here \[{a_1} = 2x\], \[{a_2} = x + 8\] and \[{a_3} = 3x + 1\].
The common difference: \[{d_1} = {a_2} - {a_1}\]
\[ \Rightarrow {d_1} = x + 8 - 2x\]
On simplification, we get
\[ \Rightarrow {d_1} = 8 - x\] ------(1)
The common difference also will be: \[{d_2} = {a_3} - {a_2}\]
\[ \Rightarrow {d_2} = 3x + 1 - \left( {x + 8} \right)\]
\[ \Rightarrow {d_2} = 3x + 1 - x - 8\]
On simplification, we get
\[ \Rightarrow {d_2} = 2x - 7\] ------(2)
As we know the common difference remains the same, then we can write
\[ \Rightarrow \,\,{d_1} = {d_2}\]
\[ \Rightarrow \,8 - x = 2x - 7\] -----(3)
Take out the variable \[x\] terms and it’s coefficients to the LHS and constant term to the RHS, then
\[ \Rightarrow \,2x + x = 8 + 7\]
\[ \Rightarrow \,3x = 15\]
Divide both side by 3, then
\[ \Rightarrow \,x = \dfrac{{15}}{3}\]
\[\therefore \,\,\,x = 5\]
Hence, the required value \[x = 5\].
Therefore, option C is the correct answer.
Note:Remember, in an arithmetic sequence, the common difference (an addition or subtraction) between any two consecutive terms of sequence is a constant or the same. We can verify the above solution by checking the common differences. Let’s consider equation (3)
\[ \Rightarrow \,8 - x = 2x - 7\]
Substitute the value \[x = 5\], then we have
\[ \Rightarrow \,8 - 5 = 2\left( 5 \right) - 7\]
\[ \Rightarrow \,8 - 5 = 10 - 7\]
\[\therefore \,\,\,3 = 3\]
Hence the common differences are the same between the two terms.
Complete step by step answer:
The general arithmetic progression is of the form \[a,a + d,a + 2d,...\] where $a$ is first term nth $d$ is the common difference which is the same between the distance on any two numbers in sequence otherwise the fixed number that must be added to any term of an AP to get the next term is known as the common difference of the AP.
The nth term of the arithmetic progression is defined as \[{a_n} = a + (n - 1)d\]. The common difference ‘\[d\]’ can be determined by subtracting the first term with the second term, second term with the third term, and so forth. Now let us consider the A.P. \[2x\], \[x + 8\], \[3x + 1\] we have to find the value of \[x\].
Here \[{a_1} = 2x\], \[{a_2} = x + 8\] and \[{a_3} = 3x + 1\].
The common difference: \[{d_1} = {a_2} - {a_1}\]
\[ \Rightarrow {d_1} = x + 8 - 2x\]
On simplification, we get
\[ \Rightarrow {d_1} = 8 - x\] ------(1)
The common difference also will be: \[{d_2} = {a_3} - {a_2}\]
\[ \Rightarrow {d_2} = 3x + 1 - \left( {x + 8} \right)\]
\[ \Rightarrow {d_2} = 3x + 1 - x - 8\]
On simplification, we get
\[ \Rightarrow {d_2} = 2x - 7\] ------(2)
As we know the common difference remains the same, then we can write
\[ \Rightarrow \,\,{d_1} = {d_2}\]
\[ \Rightarrow \,8 - x = 2x - 7\] -----(3)
Take out the variable \[x\] terms and it’s coefficients to the LHS and constant term to the RHS, then
\[ \Rightarrow \,2x + x = 8 + 7\]
\[ \Rightarrow \,3x = 15\]
Divide both side by 3, then
\[ \Rightarrow \,x = \dfrac{{15}}{3}\]
\[\therefore \,\,\,x = 5\]
Hence, the required value \[x = 5\].
Therefore, option C is the correct answer.
Note:Remember, in an arithmetic sequence, the common difference (an addition or subtraction) between any two consecutive terms of sequence is a constant or the same. We can verify the above solution by checking the common differences. Let’s consider equation (3)
\[ \Rightarrow \,8 - x = 2x - 7\]
Substitute the value \[x = 5\], then we have
\[ \Rightarrow \,8 - 5 = 2\left( 5 \right) - 7\]
\[ \Rightarrow \,8 - 5 = 10 - 7\]
\[\therefore \,\,\,3 = 3\]
Hence the common differences are the same between the two terms.
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