Question

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

Answer 1

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 $(n - k)$.