Question

# The sum of three terms which are in arithmetic progression$A.P$ is $33$ , if the product of the ${1^{st}}$ and ${3^{rd}}$ terms exceeds the ${2^{nd}}$ term by 29, find $A.P$ .

Hint: In order to solve these type of question, firstly we have to find out the $a$ and difference between two consequent terms $d$ and assuming the first three terms of $A.P$ is $\left( {a - d} \right),a,\left( {a + d} \right)$.

Assuming , second term $= a$ ,common difference$= d$
Given, That the sum of three terms is $33.$
Now,
$\left( {a - d} \right) + a + \left( {a + d} \right) = 33$
Now, opening the brackets.
Or $a - d + a + a + d = 33$
Or $3a = 33$
$a = 11 - - - - - - \left( 1 \right)$
Now, according to the given question if the product of the ${1^{st}}$ and ${3^{rd}}$ terms exceeds the ${2^{nd}}$ term by 29.
So, we have to add $29$ on $R.H.S$ or in $a$ to balance the equation .
$\left( {a - d} \right) \times \left( {a + d} \right) = a + 29$
Or ${a^2} - {d^2} = a + 29$
Substituting the value of $a$ from(1) we get
Or ${11^2} - {d^2} = 11 + 29$
Or $121 - 40 = {d^2}$
Or ${d^2} = 81$
$d = \pm 9 - - - - - \left( 2 \right)$
Due to two different values of $d$ there are two cases for arithmetic progression.
Case 1,
If $a = 11,d = 9$
$\left( {a + d} \right) = 11 + 9 = 20$
$\left( {a - d} \right) = 11 - 9 = 2$
Therefore, arithmetic progression is $2,11,20,29,38........$
Case 2,
If $a = 11,b = - 9$
$\left( {a + d} \right) = 11 + \left( { - 9} \right) = 2$
$\left( {a - d} \right) = 11 - \left( { - 9} \right) = 20$
Therefore arithmetic progression is $20,11,2, - 7, - 16..........$

Note: Whenever we face these type of question the key concept is that we know that the arithmetic progression series $\left( {a - d} \right),a,\left( {a + d} \right)$ then we have to find out the common difference of two terms using $a$ and after getting both $a$ and $d$ put their value in $A.P$ and we will get our desired answer.