Question

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

Answer 1

Hint: To find the AP we need to find the first term and the common difference. So first we find the first term and the common difference of the AP by applying the given two conditions that sum of three terms is equal to 33 and product of the 1st and 3rd term exceeds the 2nd term by 29. And finally solve these two equations to find the first term and common difference of the required AP.

Complete step-by-step answer:
Let \[\left( {a - r} \right),a,\left( {a + r} \right)\;\]be the required terms of A.P.
Therefore,

Sum of terms =33

\[\left( {a - r} \right) + a + \left( {a + r} \right) = 33\]

\[3a = 33\]

\[ \Rightarrow a = \dfrac{{33}}{3} = 11\]
Now, according to the question-
Product of 1st and 3rd term exceeds 2nd term by 29
So,
\[\begin{array}{*{20}{l}}
  {\left( {a - r} \right)\left( {a = r} \right) = a + 29} \\
  \; \\
  {{a^2} - {r^2} = a + 29} \\
  \; \\
  {{{\left( {11} \right)}^2} - {r^2} = 11 + 29} \\
  \; \\
  {121 - {r^2} = 40} \\
  \; \\
  { \Rightarrow {r^2} = 121 - 40} \\
  \; \\
  { \Rightarrow r = \pm 81 = \pm 9}
\end{array}\]

Case I:- \[r = + 9\]
\[\begin{array}{*{20}{l}}
  {a - r = 11 - 9 = 2} \\
  \; \\
  {a + r = 11 + 9 = 20}
\end{array}\]

Required A.P\[.\; = \left( {a - r} \right),a,\left( {a + r} \right) = 2,11,20\]

Case II:-
 \[\begin{array}{*{20}{l}}
  {r = - 9} \\
  \; \\
  {a - r = 11 - \left( { - 9} \right) = 20} \\
  \; \\
  {a + r = 11 + \left( { - 9} \right) = 2}
\end{array}\]

Required A.P. \[ = \left( {a - r} \right),a,\left( {a + r} \right) = 20,11,2\]

Note: Here we can suppose the three terms of AP like \[a + nd,{\text{ }}a + \left( {n + 1} \right)d,{\text{ }}a + \left( {n + 2} \right)d\] instead we supposed \[a - r,{\text{ }}a,{\text{ }}a + r\] why because that will create complex solution. Here in the second assumption the r will cancel out each other when added so we will suppose it as the second case.