Question

If Ramkali saved Rs. \[5\] in the first week of a year and then increased her weekly savings by Rs. \[1.75\]. Then, find the total number of weeks that her weekly savings become Rs. \[20.75\].
A. \[20\]
B. \[11\]
C. \[10\]
D. None of these

Answer 1

Hint: Here, we will be going to use the formula of ‘\[{t_n} = a + \left( {n - 1} \right)d\]’ which is represented in an A.P. series. As a result, considering the \[1.75\] as a common increased (difference) each week which starts from the savings of Rs. \[5\] indicates the first term of A.P. sequence and \[20.75\] is the total savings earned that is the (\[{t_n}\]) last term of the series (obtained from the given data).. Substituting all the values in the above formula and then simplifying the calculations, the desired number of weeks is/are obtained.

Complete step by step solution:
Since, it is given that
Ramkali starts savings the amount from Rs. \[5\] weekly where there is an increased of \[1.75\] per week respectively that is,
For week \[1\],
\[{a_1} = 5\]
Similarly,
For week \[2\],
\[{a_2} = 5 + 1.75 = 6.75\]
For week \[3\],
\[{a_3} = 6.75 + 1.75 = 8.50\]
.
.
.
Hence,
For \[{n^{th}}\] week,
\[{a_n} = 20.75\]
As a result, we have to find the number of weeks ‘\[n\]’ at which Ramkali will earn savings of Rs. \[20.75\]
Hence, from the above solution, it is observed that
There is the mathematical series produced,
\[ \Rightarrow 5,6.75,8.50,....,20.75\]
It seems that the series is A.P. existing the common difference ‘\[d\]’ that is,
\[d = {a_2} - {a_1} = 6.75 - 5 = 1.75\]
Hence,
To find the \[{n^{th}}\] term,
We have,
For an A.P.,
\[{t_n} = a + \left( {n - 1} \right)d\] … (i)
Where, ‘\[a\]’ is the first term in the respective series,
‘\[d\]’ is the common difference that is \[{a_2} - {a_1} = {a_3} - {a_2} = .... = {a_n} - {a_{n - 1}}\],
‘\[{t_n}\]’ is the last term of the respective series of an A.P., and
‘\[n\]’ is the total number of certain terms in the series.
Now,
Hence, substituting all the required values in the above ‘\[{t_n}\]’ formula of an A.P., we get
\[ \Rightarrow 20.75 = 5 + \left( {n - 1} \right) \times 1.75\]
Solving the equation algebraically, we get
\[ \Rightarrow 20.75 = 5 + 1.75n - 1.75\]
\[ \Rightarrow 20.75 = 3.25 + 1.75n\]
\[ \Rightarrow 20.75 - 3.25 = 1.75n\]
Hence, the number of weeks ‘\[n\]’ that the savings of Rs. \[20.75\] earned is,
\[ \Rightarrow 17.5 = 1.75n\]
\[ \Rightarrow n = \dfrac{{17.5}}{{1.75}} = \dfrac{{175 \times {{10}^{ - 1}}}}{{175 \times {{10}^{ - 2}}}}\]
Therefore, we get
\[ \Rightarrow n = \dfrac{{175}}{{175}} \times 10\]
\[ \Rightarrow n = 10weeks\]
So, the correct answer is “Option C”.

Note: One must be able to remember that the Arithmetic Progression (A.P.) series exists when there is a common difference i.e. uniform increase or decrease in the particular series/sequence (like here the difference is \[1.75\] respectively). As a result, we can use the formulae in A.P. to calculate the last term ‘\[{t_n}\]’, required number of terms ‘\[{n^{th}}\] term’ respectively by using \[{t_n} = a + \left( {n - 1} \right)d\] or the respective sum of same series by \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\], so as to be sure of our final answer.