Question

Find the $ 12th $ term from the end in A.P. $ 13,18,23,....,158. $

Answer 1

Hint: An Arithmetic Progression (AP) is the sequence of numbers in which the difference of two successive numbers is always constant.
The standard formula for Arithmetic Progression is – $ {a_n} = a + (n - 1)d $
Where $ {a_n} = $ nth term in the AP
               $ a = $ First term of AP
               $ d = $ Common difference in the series
              $ n = $ Number of terms in the AP
We are given the last term of the series, from that we will find the number of term and will count $ 12 $ from the end.

Complete step-by-step answer:
Given A.P. is $ 13,18,23,....,158. $
Here,
 $
  a = 13 \\
  d = 18 - 13 = 5 \\
  {a_n} = 158 \\
  $
Place the above values in the equation - $ {a_n} = a + (n - 1)d $
 $ \Rightarrow 158 = 13 + (n - 1)5 $
Open the bracket and simplify –
 $ \Rightarrow 158 = 13 + 5n - 5 $
Simplify the like terms on the right hand side of the equation.
\[
   \Rightarrow 158 = \underline {13 - 5} + 5n \\
   \Rightarrow 158 = 8 + 5n \\
 \]
When the number is moved from one side to other, the sign also changes from positive to negative and vice-versa.
 $
   \Rightarrow 158 - 8 = 5n \\
   \Rightarrow 150 = 5n \\
  $
When the term in multiplicative changes its side, it goes in division.
 $
   \Rightarrow n = \dfrac{{150}}{5} \\
   \Rightarrow n = 30 \\
  $
Now, the $ 12th $ term from end $ = n - 12 + 1 $
 $ 12th $ Term from end $ = 30 - 12 + 1 $
 $ 12th $ Term from end $ = 19th $ from the beginning
Now, $ 19th $ term is –
 $ {a_{19}} = a + (19 - 1)d $
Place the values of “a” and “d” and simplify –
 $ {a_{19}} = 13 + (18)(5) $
Simplify –
 $
  {a_{19}} = 13 + 90 \\
  {a_{19}} = 103 \\
  $
Hence the $ 12th $ term from the end in A.P. $ 13,18,23,....,158 $ is $ 103 $

Note: Know the difference between the geometric progression and the arithmetic progression and apply accordingly. There are two types of sequences and series.
Arithmetic progression
Geometric Progression.
In arithmetic progression, the difference between the numbers is constant in the series whereas the geometric progression is the sequence in which the succeeding element is obtained by multiplying the preceding number by the constant and the same continues for the series. The ratio between the two remains the same.
The arithmetic series is the sum of all the terms of the arithmetic progression (AP). Where, “a” is the first term and “d” is the common difference among the series.
 $
  {S_n}{\text{ = a + (a + d) + (a + 2d) + (a + 3d) + }}......{\text{ + [a + (n - 1)d]}} \\
  {{\text{S}}_{n{\text{ }}}} = \dfrac{n}{2}[2a + (n - 1)d] \\
  $