Question

The sum of the first $9$ terms of an AP is $81$ and the sum of its first $20$ terms is $400$. Find the first term and the common difference of the AP.

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 –
${S_n} = \dfrac{n}{2}[2a + (n - 1)d]$
Where ${S_n} = $ Sum of n terms in AP
              $a = $ First term of AP
              $d = $ Common difference in the series
             $n = $ Number of terms in the AP

Complete step by step solution:
Given that: The sum of the first $9$ terms of an AP is $81$
Put, the known values in the standard equation –
${S_n} = \dfrac{n}{2}[2a + (n - 1)d]$
$81 = \dfrac{9}{2}[2a + (9 - 1)d]$
Simplifying the above equation –
$162 = 9[2a + (8)d]$
$\dfrac{{169}}{9} = [2a + 8d]$
$18 = 2(a + 4d)$
$9 = a + 4d$ ................................(A)
Now, the second condition given-
The sum of its first $20$ terms is $400$.
Substituting the known values –
${S_n} = \dfrac{n}{2}[2a + (n - 1)d]$
$400 = \dfrac{{20}}{2}[2a + (20 - 1)d]$
Simplifying the above equations –
$800 = 20[2a + 19d]$
$40 = 2a + 19d$ .................................(B)
Taking both the equations together –
 $9 = a + 4d$
 $40 = 2a + 19d$
To use elimination methods -
Multiplying the equation with $2$ and re-writing it-
$18 = 2a + 8d$ ...................(C)
$40 = 2a + 19d$ ..................(B)
Subtracting the Equation (B) – (C)
$ \Rightarrow 22 = 11d$
$d = \dfrac{{22}}{{11}}$
$ \Rightarrow d = 2$
Substitute the values of $d = 2$, in Equation (A)
$9 = a + 4d$
$9 = a + 4(2)$
Taking “a” subject –
$a = 9 - 8$
$a = 1$
Therefore, the required solution are-
The first term of the arithmetic progression is $a = 1$
And the common difference of the arithmetic progression is $d = 2$.
Note: Instead of the elimination method, you can use substitution method to solve the equations and to find the unknown quantities