Question

Find the correct option
If \[{A_1},{A_2}\] are two arithmetic means between \[\dfrac{1}{3}\] and \[\dfrac{1}{{24}}\], then their values are
A. \[\dfrac{7}{{72}},\dfrac{5}{{36}}\]
B. \[\dfrac{{17}}{{72}},\dfrac{5}{{36}}\]
C. \[\dfrac{7}{{36}},\dfrac{5}{{72}}\]
D. \[\dfrac{5}{{72}},\dfrac{{17}}{{72}}\]

Answer 1

Hint: Using the formula for the \[n\]-th term of an arithmetic progression and assuming the common difference as a variable, an equation will be established with the given numerical value of the last i.e. the fourth term of the arithmetic progression formed by the four numbers. Then, the numerical value of the common difference will be calculated to find the two arithmetic means i.e. the second term and the third term of the arithmetic progression.
Formulae Used: The \[n\]-th term of an arithmetic progression is represented by the formula
\[{t_n} = a + (n - 1)d\], where,
\[{t_n} = \] the \[n\]-th term of the A.P.
\[a = \] the first term of the A.P.
\[d = \] the common difference of the A.P.
\[n = \] the number of terms of the A.P.

Complete step-by-step solution:
We have been given that \[{A_1},{A_2}\] are two arithmetic means between \[\dfrac{1}{3}\] and \[\dfrac{1}{{24}}\].
We have to find out the above two arithmetic means.
The above four numbers i.e. \[\dfrac{1}{3},{A_1},{A_2},\dfrac{1}{{24}}\] are in a an arithmetic progression, in which,
The first term \[ = \dfrac{1}{3}\]
The last term i.e. the fourth term \[ = \dfrac{1}{{24}}\]
The number of terms \[ = 4\]
Let, the common difference \[ = d\]
Assigning the values to the variables of the formulae, we have
\[{t_4} = \dfrac{1}{{24}}\]
\[a = \dfrac{1}{3}\]
\[n = 4\]
Substituting the values in the corresponding formula \[{t_n} = a + (n - 1)d\], we have
\[{t_4} = \dfrac{1}{3} + (4 - 1)d\]
\[ \Rightarrow \dfrac{1}{{24}} = \dfrac{1}{3} + 3d\]
\[ \Rightarrow 3d = - \dfrac{7}{{24}}\]
\[ \Rightarrow d = - \dfrac{7}{{72}}\]
Applying the above formula, the second and third terms of the A.P. i.e. the two arithmetic means \[{A_1},{A_2}\] are calculated.
The second term \[ = \] First arithmetic mean i.e. \[{t_2} = {A_1}\]
Taking \[n\] equal to \[2\], we have
\[{t_2} = a + (2 - 1)d\]
\[ = \dfrac{1}{3} + ( - \dfrac{7}{{72}})\]
\[ = \dfrac{{17}}{{72}}\]
So, \[{A_1} = \dfrac{{17}}{{72}}\]
Similarly, the third term \[ = \] Second arithmetic mean i.e. \[{t_3} = {A_2}\]
Taking \[n\] equal to \[3\], we have
\[{t_3} = a + (3 - 1)d\]
\[ = \dfrac{1}{3} + 2 \times ( - \dfrac{7}{{72}})\]
\[ = \dfrac{1}{3} - \dfrac{7}{{36}}\]
\[ = \dfrac{5}{{36}}\]
So, \[{A_2} = \dfrac{5}{{36}}\]

Hence, option B. \[\left( {\dfrac{{17}}{{72}},\dfrac{5}{{36}}} \right)\] is the correct answer.

Note: The formula \[{t_n} = a + (n - 1)d\] is used to find the \[n\]-th term \[{t_n}\] of an A.P. with \[a\] and \[d\] as first term and common difference respectively. The common difference in an A.P. can be calculated by subtracting its any term from the succeeding term. This formula is not applicable for a G.P.