Question

Find the ${60^{th}}$ term of AP 8, 10, 12,………, If it has a total of 60 terms and hence find the sum of its last 10 terms.

Answer 1

Hint: To solve the above arithmetic progression question, first we will obtain the ${60^{th}}$ term and then ${51^{th}}$ term using formula ${a_n} = a + (n - 1)d$ and then we will obtain the sum of the last 10 terms of the AP using the formula $S_n = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$

Complete step-by-step answer:
The given arithmetic progression is 8, 10, 12,………….
Here the first term $a = 8$
Common difference, $d = 10 - 8 = 2$
For the ${60^{th}}$ term of AP, $n = 60$
Hence the ${60^{th}}$term of the given AP is,
${a_n} = a + (n - 1)d$
Putting the values of a, n and d in the given formulae we get,
${a_{60}} = 8 + (60 - 1)2$
$ \Rightarrow {a_{60}} = 8 + 59 \times 2$
$ \Rightarrow {a_{60}} = 126$
Hence the ${60^{th}}$ term of the given AP is 126.
Let the last 10 terms of the given AP is ${a_{51}},{a_{52}},{a_{53}},..............,{a_{60}}$
Using above formula we will get ${51^{th}}$
Hence, ${a_{51}} = 8 + (51 - 1)2$
$ \Rightarrow {a_{51}} = 8 + 100 = 108$
Let, ${a_{51}},{a_{52}},{a_{53}},..............,{a_{60}}$is another AP having $a = 108,d = 2$
Hence the sum of the 10 terms of the AP is given by,
$S_n = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
Putting the values of n, a and d in the above formula we get,
$S_10 = \dfrac{{10}}{2}\left[ {2 \times 108 + \left( {10 - 1} \right)2} \right]$
Simplifying the above equation we get,
$S_10 = 5\left[ {216 + 18} \right]$
Or, $S_10 = 5 \times 234 = 1170$
Therefore, the ${60^{th}}$term of the given AP is 126 and the sum of its last 10 terms is 1170.

Note: An arithmetic progression is a sequence of numbers such that the difference of any two consecutive numbers is constant.
The formula for the $n^{th}$ term of the arithmetic progression is given by,
${a_n} = a + (n - 1)d$
Where ${a_n}$is the nth term in the sequence, a is the first term of the sequence and d is the common difference.
The formula for the sum of the n terms of the AP is,
$S_n = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
Where n is number of terms, a is the first term if the sequence to be added and d is the common difference.
You should remember all the formulas of arithmetic progression and geometric progression.