Question

Find the number of terms in an A.P if, the sum of three consecutive terms of the A.P is 24 and their product is 312? (The term of A.P are in ascending order)

Answer 1

Hint:Take the three consecutive terms of the A.P as $a - d,a{\text{ and }}a + d$.
The given conditions are:
sum of three consecutive terms is = 24
$ \Rightarrow \left( {a - d} \right) + a + \left( {a + d} \right) = 24$
Also, Product of three consecutive terms = 312
$ \Rightarrow \left( {a - d} \right) \times a \times \left( {a + d} \right) = 312$
 Solve the above equations simultaneously to find a and d, after this you can find all the consecutive terms easily.

Complete step-by-step solution:
Given,
The sum of three consecutive terms is = 24
Product of three consecutive terms = 312
In A.P (arithmetic progression) the difference between two consecutive terms is always constant.
Generally, A.P is described as:
 $a,a + d,a + 2d,a + 3d..............{\text{ and so on}}{\text{.}}$
Where a is the beginning term or first term and d is the common difference between any two consecutive terms.
Let the three consecutive terms of the A.P be $a - d,a{\text{ and }}a + d$
Since, sum of three consecutive terms is = 24
$
   \Rightarrow \left( {a - d} \right) + a + \left( {a + d} \right) = 24 \\
   \Rightarrow 3a = 24 \\
  \therefore a = 8 \\
$
Also, Product of three consecutive terms = 312
$
   \Rightarrow \left( {a - d} \right) \times a \times \left( {a + d} \right) = 312 \\
   \Rightarrow \left( {8 - d} \right) \times 8 \times \left( {8 + d} \right) = 312{\text{ }}\left( {\because a = 8} \right) \\
   \Rightarrow \left( {8 - d} \right) \times \left( {8 + d} \right) = \dfrac{{312}}{8} \\
   \Rightarrow 64 - {d^2} = 39{\text{ }}\left( {\because \left( {a - b} \right) \times \left( {a + b} \right) = {a^2} - {b^2}} \right) \\
   \Rightarrow {d^2} = 64 - 39 \\
   \Rightarrow {d^2} = 25 \\
   \Rightarrow d = \pm 5 \\
$
Neglecting d = -5 because in question it is given terms of A.P are in ascending order but on taking d = -5 we get an A.P whose terms are in descending order.
Taking a = 8 and d = 5 we find the terms of A.P:
$
  a - d = 8 - 5 = 3 \\
  a = 8 \\
  a + d = 8 + 5 = 13 \\
$
The three consecutive terms of the given A.P are 3, 8 and 13

Additional information: Sometimes in many questions we have to assume a certain number of terms in Arithmetic Progression. The following methods are generally used for the selection of terms in an arithmetic progression:
>If the sum of the three consecutive terms in A.P is given then, we take the numbers to be\[\;a{\text{ }} - {\text{ }}d,{\text{ }}a{\text{ }}and{\text{ }}a{\text{ }} + {\text{ }}d\]. Here, the common difference is\[d\].
>If the sum of four consecutive terms in A.P is given then, we take the numbers to be \[a{\text{ }} - {\text{ }}3d,{\text{ }}a{\text{ }} - {\text{ }}d,{\text{ }}a{\text{ }} + {\text{ }}d{\text{ }}and{\text{ }}a{\text{ }} + {\text{ }}3d\]. Here the common difference is\[2d\].
>If the sum of five consecutive terms in A.P is given then, we take the numbers to be\[a{\text{ }} - {\text{ }}2d,{\text{ }}a{\text{ }} - {\text{ }}d,{\text{ }}a,{\text{ }}a{\text{ }} + {\text{ }}d{\text{ }}and{\text{ }}a{\text{ }} + {\text{ }}2d\]. Here the common difference is\[d\].
>If the sum of six consecutive terms in A.P is given then, we take the numbers to be\[a{\text{ }} - {\text{ }}5d,{\text{ }}a{\text{ }} - {\text{ }}3d,{\text{ }}a{\text{ }} - {\text{ }}d,{\text{ }}a{\text{ }} + {\text{ }}d,{\text{ }}a{\text{ }} + {\text{ }}3d{\text{ }}and{\text{ }}a{\text{ }} + {\text{ }}5d\]. Here the common difference is\[2d\] .

Note:From the above explanation, we can note that in case of an odd number of consecutive terms, the middle term will be a and the common difference will be d.
And in case of an even number of consecutive terms the middle terms are \[a{\text{ }} - {\text{ }}d,{\text{ }}a{\text{ }} + {\text{ }}d\]and the common difference is 2d.