Question

Find the sum of last $10$ terms of an A.P. $8,10,12,.....................,126$.

Answer 1

Hint: Here we have a series of terms in A.P. To find the sum of last $10$ terms, we need to reverse the given A.P., because the sum of last ten terms of the AP: $8,10,12,.....................,126$ is the same as the sum of the first ten terms of the AP: $126,124,122,...................,12,10,8$.

Complete step-by-step answer:
Given AP is: $8,10,12,.....................,126$
Reversing all terms of given AP, we get another AP, i.e., $126,124,122,...................,12,10,8$.
Here, first term $\left( a \right) = 126$
& Common difference $\left( d \right) = 124 - 126 = - 2$
The sum of $n$ terms of an AP is given by,
${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
To find the sum of $10$ terms of AP, put the values of $a$ and $d$ and $n = 10$, we get-
${S_{10}} = \dfrac{{10}}{2}\left[ {2 \times 126 + \left( {10 - 1} \right)\left( { - 2} \right)} \right]$
$ \Rightarrow {S_{10}} = 5\left[ {252 - 18} \right]$
$ \Rightarrow {S_{10}} = 5 \times 234$
$ \Rightarrow {S_{10}} = 1170$

Hence the sum of last $10$ terms of the A.P. $8,10,12,.....................,126$ is $1170$.

Note: An another approach to solve this question is described below:
Given AP is: $8,10,12,.....................,126$
Here, first term $\left( a \right) = 8$
Last term $\left( {{a_n}} \right) = 126$
& Common difference $\left( d \right) = 10 - 8 = 2$
The $n$th term of an AP is given by,
${a_n} = a + \left( {n - 1} \right)d$
For finding number of terms $\left( n \right)$, put the values of $a$ and $d$ and${a_n} = 126$, we get-
$126 = 8 + \left( {n - 1} \right)\left( 2 \right)$
$126 = 8 + 2n - 2$
$126 = 6 + 2n$
$126 - 6 = 2n$
$n = \dfrac{{120}}{2}$
$n = 60$
Hence, the given AP has total $60$ terms.
Now, Sum of last $10$ terms of the given AP= Sum of first $60$ terms $ - $ Sum of first $50$ terms
The sum of $n$ terms of an AP is given by,
${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
$\therefore $ Sum of last $10$ terms of the given AP $ = {S_{60}} - {S_{50}}$
$ = \dfrac{{60}}{2}\left[ {2 \times 8 + \left( {60 - 1} \right)\left( 2 \right)} \right] - \dfrac{{50}}{2}\left[ {2 \times 8 + \left( {50 - 1} \right)\left( 2 \right)} \right]$
$ = 30\left[ {16 + 118} \right] - 25\left[ {16 + 98} \right]$
$ = 30 \times 134 - 25 \times 114$
$ = 4020 - 2850$
$ = 1170$
Hence the sum of last $10$ terms of the A.P. $8,10,12,.....................,126$ is $1170$.