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Question: The sum of the first three terms of an AP is \[33\]. If the product of the first and the third excee...

The sum of the first three terms of an AP is 3333. If the product of the first and the third exceeds the second term by 2929, find the AP.

Explanation

Solution

Hint: We know that Arithmetic Progression is a sequence of numbers in which the next term is obtained by adding a certain number to the previous term. We will take the second term to be aaand then find the first term and the third term respectively using the common difference dd.

Complete step-by-step answer:
The question tells us that the sum of the first three terms is 3333 and that the product of the first and the third term exceeds the second term by 2929.
So, let us assume that the second term of this AP to be aa and the common difference between the terms to be dd.
If the second term is aa, then the first term will be ada - d and the third term will be a+da + d.
So, according to the question, we get,
(ad)+a+(a+d)=33(a - d) + a + (a + d) = 33 and
(ad)(a+d)=a+29(a - d)(a + d) = a + 29
When we solve (ad)+a+(a+d)=33(a - d) + a + (a + d) = 33, we get

ad+a+a+d=33 3a=33 a=11  \Rightarrow a - d + a + a + d = 33 \\\ \Rightarrow 3a = 33 \\\ \Rightarrow a = 11 \\\

When we solve (ad)(a+d)=a+29(a - d)(a + d) = a + 29, we get

(ad)(a+d)=a+29 a2d2=a+29  (a - d)(a + d) = a + 29 \\\ \Rightarrow {a^2} - {d^2} = a + 29 \\\

Now, we will substitute a=11a = 11in the above equation,

a2d2=a+29 112d2=11+29 121d2=40 12140=d2 d2=81 d=±81 d=±9  {a^2} - {d^2} = a + 29 \\\ \Rightarrow {11^2} - {d^2} = 11 + 29 \\\ \Rightarrow 121 - {d^2} = 40 \\\ \Rightarrow 121 - 40 = {d^2} \\\ \Rightarrow {d^2} = 81 \\\ \Rightarrow d = \pm \sqrt {81} \\\ \Rightarrow d = \pm 9 \\\

Either d=9d = 9 or d=9d = - 9
When a=11a = 11and d=9d = 9,
The first term=ad=119=2 = a - d = 11 - 9 = 2
The third term=a+d=11+9=20 = a + d = 11 + 9 = 20
Therefore, the AP series: 2,11,20,29,....2,11,20,29,....
When a=11a = 11 and d=9d = - 9,
The first term=ad=11(9)=20 = a - d = 11 - ( - 9) = 20
The third term=a+d=11+(9)=119=2 = a + d = 11 + ( - 9) = 11 - 9 = 2
Therefore, the AP series: 20,11,2,...20,11,2,...
Thus, the two AP series are 2,11,20,29,....2,11,20,29,.... and 20,11,2,...20,11,2,....

Note: In these types of questions where the terms of the sequence are not given and have to be assumed, we might make the mistake of assuming the terms as separate variables like a, b, c and so on. We need to remember that whenever we are given the sum of three consecutive terms, we start the terms with ada - d.