Question

How many terms of the AP \[20\], \[19\dfrac{1}{3}\], \[18\dfrac{2}{3}\],… must be taken to make the sum \[300\]. Explain the answer.

Answer 1

Hint: Here we will use the formula of summation of AP series which states as below:
\[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\] , where \[{S_n}\] is the sum of \[n^{th}\] terms in AP series, \[a\] is equals to the first term of the series and \[d\] is the common difference between two terms of the series.

Complete step-by-step answer:
Step 1: For the given AP series, \[20\], \[19\dfrac{1}{3}\], \[18\dfrac{2}{3}\],… we need to calculate the \[n^{th}\] term at which the sum will be equals to \[300\]. By using the formula for summation, we get:
\[ \Rightarrow {S_n} = \dfrac{n}{2}\left[ {2\left( {20} \right) + \left( {n - 1} \right)\left( { - \dfrac{2}{3}} \right)} \right]\] , where \[a = 20\] and \[d\] will be equal to the difference between the two terms of the series as shown below:
\[ \Rightarrow d = \left( {19\dfrac{1}{3} - 20} \right)\]
After doing the subtraction inside the bracket we get:
\[ \Rightarrow d = - \dfrac{2}{3}\]
Step 2: As given in the question, that sum should be equal to \[300\]. By substituting this value in the equation \[{S_n} = \dfrac{n}{2}\left[ {2\left( {20} \right) + \left( {n - 1} \right)\left( { - \dfrac{2}{3}} \right)} \right]\], we get:
\[ \Rightarrow 300 = \dfrac{n}{2}\left[ {2\left( {20} \right) + \left( {n - 1} \right)\left( { - \dfrac{2}{3}} \right)} \right]\] ……………… (1)
By opening the brackets in the RHS side of the equation (1), we get:
\[ \Rightarrow 300 = \dfrac{n}{2}\left[ {40 - \dfrac{2}{3}n + \dfrac{2}{3}} \right]\] ………… (2)
By taking LCM inside the brackets into the RHS side of the equation (2), we get:
\[ \Rightarrow 300 = \dfrac{n}{2}\left[ {\dfrac{{120 - 2n + 2}}{3}} \right]\]
By doing simple addition inside the brackets into the RHS side, we get:
\[ \Rightarrow 300 = \dfrac{n}{2}\left[ {\dfrac{{122 - 2n}}{3}} \right]\]
By opening the brackets and doing simple multiplication into the RHS side of the equation \[300 = \dfrac{n}{2}\left[ {\dfrac{{122 - 2n}}{3}} \right]\], we get:
\[ \Rightarrow 300 = \dfrac{{122n - 2{n^2}}}{6}\]
By taking \[6\] into the LHS side and after multiplying it with \[300\], we get:
\[ \Rightarrow 1800 = 122n - 2{n^2}\]
By dividing the complete equation \[1800 = 122n - 2{n^2}\] with \[2\]:
\[ \Rightarrow 900 = 61n - {n^2}\]
By bringing the whole terms at one side for making it a perfect quadratic equation, we get:
\[ \Rightarrow {n^2} - 61n + 900 = 0\] ……………… (3)
Step 3: In equation (3), for finding the value of \[n\], we will use the factorization method by using which we will find the terms whose product will be equals to \[900\] and addition or subtraction will be equals to \[61\] as shown below:
\[ \Rightarrow {n^2} - 36n - 25n + 900 = 0\]
By taking \[n\]common from the first two terms and \[25\] common from the last two terms in the equation \[{n^2} - 36n - 25n + 900 = 0\], we get:
\[ \Rightarrow n\left( {n - 36} \right) - 25\left( {n - 36} \right) = 0\]
By taking \[\left( {n - 36} \right)\] common we get:
\[ \Rightarrow \left( {n - 25} \right)\left( {n - 36} \right) = 0\] ………… (4)
Now, from equation (4), we have two values of \[n\] as shown below:
\[ \Rightarrow n = 25,36\]
 Sum of the AP series till \[25th\] and \[36th\] term will be equal to \[300\].
\[ \Rightarrow {S_{25th}} = 300\], \[{S_{36th}} = 300\]

Sum of the AP series till \[25^{th}\] and \[36^{th}\] term will be equal to \[300\].

Note: Students need to remember the formula for AP and GP series whose full forms are Arithmetic Progression and Geometric progression series.
Also, you should take care that if the last term of the AP series is given then the formula for finding the sum will be different as shown below:
\[ \Rightarrow {S_n} = \dfrac{n}{2}\left( {a + l} \right)\] , \[n = nth{\text{ term}}\], \[a\] equal to the first term of the series and \[l\] equals to the last terms of the series.