Question

Find the \[{20^{th}}\] term from the last of the AP: \[3,8,13,.....,253\].

Answer 1

Hint: For finding the \[{20^{th}}\] term from the last of the AP: \[3,8,13,.....,253\], we will first find the number of terms in the AP using the formula \[l = a + (n - 1)d\], where
\[l = \] last term of the AP
\[n = \] number of terms in the AP
\[a = \] first term of the AP
\[d = \] common differences between the terms
Now, after finding the number of terms in the AP, we will find out which term from the starting will be the \[{20^{th}}\] term from the last using the formula, \[{k^{th}}\] term from the last \[ = {\left( {n - (k - 1)} \right)^{th}}\] term from the starting , say \[{a_m}\], where
\[{a_m} = \] \[{m^{th}}\] term from the starting
\[n = \] total number of terms in the AP
After that, we will find the \[{a_m}^{th}\] term using the formula \[{a_m} = a + (m - 1)d\].

Complete answer: We are given AP: \[3,8,13,......,253\].
Here, \[a = {a_1} = 3\]
\[d = {a_2} - {a_1} = 8 - 3 = 5\]
\[l = 253\]
Let the total number of terms be
Now, Finding the total number of terms using the formula \[l = a + (n - 1)d\]
Substituting the values, we have
\[253 = 3 + (n - 1)5\]
Rearranging the terms, we get
\[ \Rightarrow 253 - 3 = (n - 1)5\]
\[ \Rightarrow 250 = (n - 1)5\]
Now, dividing the equation by \[5\], we get
\[ \Rightarrow \dfrac{{250}}{5} = \dfrac{{(n - 1)5}}{5}\]
Cancelling out the terms, we get
\[ \Rightarrow 50 = (n - 1)\]
Rearranging the terms again, we get
\[ \Rightarrow 50 + 1 = n\]
\[ \Rightarrow 51 = n - - - - - - (1)\]
Hence, we get the total number of terms to be equal to \[51\]
Now, \[{20^{th}}\]term from the last = \[{\left( {n - (20 - 1)} \right)^{th}}\] term from the starting
Now, using (1), we get
Now, \[{20^{th}}\]term from the last = \[{\left( {n - (20 - 1)} \right)^{th}}\] term from the starting
\[ = {\left( {51 - \left( {20 - 1} \right)} \right)^{th}}\] term from the starting
\[ = {\left( {51 - 19} \right)^{th}}\]term from the starting
\[ = {32^{nd}}\] term from the starting
Hence, we get,
\[{20^{th}}\]term from the last \[ = {32^{nd}}\] term from the starting
\[{20^{th}}\]term from the last\[ = {a_{32}}\]
Now, finding the value of \[{a_{32}}\] using \[{a_m} = a + (m - 1)d\]
We have,
\[{a_{32}} = a + (32 - 1)d\]
We have \[a = 3\] and \[d = 5\]. Using this in above formula, we get
\[{a_{32}} = 3 + (32 - 1)5\]
\[ \Rightarrow {a_{32}} = 3 + (31)5\]
Solving further, we get
\[ \Rightarrow {a_{32}} = 3 + 155\]
\[ \Rightarrow {a_{32}} = 158\]
Hence, we get, \[{20^{th}}\]term from the last of AP: \[3,8,13,.....,253\] is \[158\].

Note:
We could have used another method of solving this problem. As we have to find the \[{20^{th}}\] term from the last of AP: \[3,8,13,.....,253\], we could have written the AP in reverse order and then find the \[{20^{th}}\] term from the starting of the AP written in reverse order. While using the formula we have used above, When we are finding the \[{k^{th}}\] from the last as the \[{m^{th}}\] term from the beginning, we should remember that the formula for that is \[(n - (k - 1))\] not just \[(n - k)\].