Question

If the sum and product of the first three terms in an AP are 33 and 1155 respectively, then value of its ${{11}^{th}}$ term is:
a. -25
b. 25
c. -36
d. 35

Answer 1

In this question, we have been given that the sum and product of the first three terms of an AP are 33 and 1155 respectively. So, we will consider the three terms in AP as (a-d), a, (a+d), where (a-d) is the first term, ‘a’ the second term and (a+d) the third term. Then we will get the value of ‘a’ from the equation of sum of the three numbers and the value of ‘d’ from the equation of the product of the three numbers. Then finally, we will use the general term formula, ${{t}_{n}}=a+\left( n-1 \right)d$ to find the ${{11}^{th}}$ term.

Complete step-by-step answer:
The question given to us is related to the AP, or arithmetic progression. So, let us understand what an AP is. An AP is a sequence of numbers such that the difference of any two successive numbers is a constant.
From the given data, we have the sum of the three terms in AP as 33 and the product of those three numbers as 1155. Now, let us consider the 3 terms in AP as (a-d), a, (a+d).
So, the sum of these three numbers can be equated to 33, so we can write,
$\left( a-d \right)+a+\left( a+d \right)=33$
On solving further, we get,
$3a=33$
On dividing the above equation by 3, we get,
$a=\dfrac{33}{3}=11$
Hence, we get the value of a as 11.
Now, we can equate the product of the three numbers to 1155, so we get,
$\left( a-d \right)\left( a \right)\left( a+d \right)=1155$
Now, we know that $\left( x+y \right)\left( x-y \right)={{x}^{2}}-{{y}^{2}}$, so we can write the above equation as,
$a\left( {{a}^{2}}-{{d}^{2}} \right)=1155.........(i)$
So, on substituting the value of a = 11 in equation (i), we get,
$11\left( {{11}^{2}}-{{d}^{2}} \right)=1155$
On dividing the above equation by 11 on both the sides, we get,
$\left( {{11}^{2}}-{{d}^{2}} \right)=\dfrac{1155}{11}$
On solving further, we get,
$\begin{aligned} & 121-{{d}^{2}}=105 \\\ & {{d}^{2}}=121-105 \\\ & {{d}^{2}}=16 \\\ & d=\pm 4 \\\ \end{aligned}$
So, we have two values for d, that is d = 4 and d = -4.
Considering d = 4, we get the three numbers, $\left( a-d \right),\left( a \right),\left( a+d \right)$ as $\left( 11-4 \right),\left( 11 \right),\left( 11+4 \right)\Rightarrow 7,11,15$. So, the three numbers in AP are 7, 11 and 15. Hence, we get the first term of this AP as 7.
Considering d = -4, we get the three numbers, $\left( a-d \right),\left( a \right),\left( a+d \right)$ as $\left( 11-\left( -4 \right) \right),\left( 11 \right),\left( 11+\left( -4 \right) \right)\Rightarrow 15,11,7$. So, the three numbers in AP are 15, 11 and 7. Hence, we get the first term of this AP as 15.
Now, we are asked to find the ${{11}^{th}}$ term, so we can apply the general term formula, that is, ${{t}_{n}}=a+\left( n-1 \right)d$.
Here, ‘a’ represents the first term of the AP, ‘d’ represents the common difference and ‘n’ represents the number of the term.
First, we will find the ${{11}^{th}}$ term when d = 4 and a = 7, so on substituting n = 11, a = 7, d = 4 in the general term formula we get,
$\begin{aligned} & {{t}_{11}}=7+\left( 11-1 \right)\left( 4 \right) \\\ & {{t}_{11}}=7+\left( 10 \right)\left( 4 \right) \\\ & {{t}_{11}}=7+40 \\\ & {{t}_{11}}=47 \\\ \end{aligned}$
Now, we will find the ${{11}^{th}}$ term when d = -4 and a = 15, so on substituting n = 11, a = 15, d = -4 in the general term formula we get,
$\begin{aligned} & {{t}_{11}}=15+\left( 11-1 \right)\left( -4 \right) \\\ & {{t}_{11}}=15+\left( 10 \right)\left( -4 \right) \\\ & {{t}_{11}}=15-40 \\\ & {{t}_{11}}=-25 \\\ \end{aligned}$
Therefore, we get the values of the ${{11}^{th}}$ term as 47 and -25.

So, the correct answer is “Option (a)”.

Note: The most common mistake that the students can make while solving this question is that, if they assume the three terms of the AP as (a-d), a, (a+d), in the final step, they may consider the first term of the AP as ‘a’ and not (a-d) and this will lead to errors in the final answer. Also, an alternate way to solve this question would be that, the students can assume the three terms of the AP as a, (a+d), (a+2d) and then proceed with the solution.