The four arithmetic mean between 3 and 23 are:-
A. 5,9,11,13
B. 7,11,15,19
C. 7,11,15,22
D. 7,15,19,21
Answer
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Hint: This question can be done by easily understanding the concepts of arithmetic progression which are mentioned below.
*$n^{th}$ term of an A.P (arithmetic progression) is given by ${T_n} = a + (n - 1)d$ where a= first term of the sequence and d=common difference which is given by $d = {T_n} - {T_{n - 1}}$. We will substitute the value of ‘a’ in the sixth term of the arithmetic sequence, we get the value of ‘d’. Then we will find the four terms of sequence just by substituting the value of ‘a’, ‘d’ and ‘n’ in the general equation.
Complete step-by-step answer:
Let four arithmetic means be ${T_2},{T_3},{T_4},{T_5}$ between 3 and 23. Then arithmetic progression is given by $3, {T_2}, {T_3}, {T_4}, {T_5}, 23$
Now we will apply formula ${T_n} = a + (n - 1)d$ so that we can find out the sixth term of the sequence.
$ \Rightarrow {T_6} = 3 + (6 - 1)d$ (where a=3 and n=6)
$ \Rightarrow 23 = 3 + (5)d$ (Sixth term is 23)
$ \Rightarrow 5d = 23 - 3$
$ \Rightarrow 5d = 20$
$\therefore d = \dfrac{{20}}{5} = 4$
Therefore common difference is 4
Now we will apply formula $d = {T_n} - {T_{n - 1}}$ which can also be written as \[{T_n} = d + {T_{n - 1}}\]
\[ \Rightarrow {T_2} = 4 + {T_{2 - 1}}\]
\[ \Rightarrow {T_2} = 4 + {T_1}\]
\[\therefore {T_2} = 4 + 3 = 7\]
Now from the above process we can clearly see that the succeeding term can be found by adding the common difference of the proceeding term.
\[\therefore {T_3} = 4 + 7 = 11\]
\[\therefore {T_4} = 4 + 11 = 15\]
\[\therefore {T_4} = 4 + 15 = 19\]
Therefore final answer i.e. (7, 11, 15, 19)
So, the correct answer is “Option B”.
Note: Alternate method: - Students can also do this question by directly observing the options. As we know that in arithmetic sequence common difference is difference between the consecutive terms and is constant value for the entire sequence. So as we see in option (A) the common difference is not constant same goes with option (C) and option (D) only in option (B) the common difference is constant so it will be the correct answer. This method only satisfies here because other options were eliminated but proper method should be followed which is mentioned earlier.
*$n^{th}$ term of an A.P (arithmetic progression) is given by ${T_n} = a + (n - 1)d$ where a= first term of the sequence and d=common difference which is given by $d = {T_n} - {T_{n - 1}}$. We will substitute the value of ‘a’ in the sixth term of the arithmetic sequence, we get the value of ‘d’. Then we will find the four terms of sequence just by substituting the value of ‘a’, ‘d’ and ‘n’ in the general equation.
Complete step-by-step answer:
Let four arithmetic means be ${T_2},{T_3},{T_4},{T_5}$ between 3 and 23. Then arithmetic progression is given by $3, {T_2}, {T_3}, {T_4}, {T_5}, 23$
Now we will apply formula ${T_n} = a + (n - 1)d$ so that we can find out the sixth term of the sequence.
$ \Rightarrow {T_6} = 3 + (6 - 1)d$ (where a=3 and n=6)
$ \Rightarrow 23 = 3 + (5)d$ (Sixth term is 23)
$ \Rightarrow 5d = 23 - 3$
$ \Rightarrow 5d = 20$
$\therefore d = \dfrac{{20}}{5} = 4$
Therefore common difference is 4
Now we will apply formula $d = {T_n} - {T_{n - 1}}$ which can also be written as \[{T_n} = d + {T_{n - 1}}\]
\[ \Rightarrow {T_2} = 4 + {T_{2 - 1}}\]
\[ \Rightarrow {T_2} = 4 + {T_1}\]
\[\therefore {T_2} = 4 + 3 = 7\]
Now from the above process we can clearly see that the succeeding term can be found by adding the common difference of the proceeding term.
\[\therefore {T_3} = 4 + 7 = 11\]
\[\therefore {T_4} = 4 + 11 = 15\]
\[\therefore {T_4} = 4 + 15 = 19\]
Therefore final answer i.e. (7, 11, 15, 19)
So, the correct answer is “Option B”.
Note: Alternate method: - Students can also do this question by directly observing the options. As we know that in arithmetic sequence common difference is difference between the consecutive terms and is constant value for the entire sequence. So as we see in option (A) the common difference is not constant same goes with option (C) and option (D) only in option (B) the common difference is constant so it will be the correct answer. This method only satisfies here because other options were eliminated but proper method should be followed which is mentioned earlier.
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