Three numbers square measure during A.P. whose add is thirty-three and products is 792, then the tiniest variety from these numbers is
A) 4
B) 8
C) 11
D) 14
Answer
298.2k+ views
Hint: If the three unknown numbers are given which are in arithmetic progression (A.P), then unknown numbers of arithmetic series are always written as (a – d), a, (a + d). After that apply the conditions on these numbers as mentioned in the question.
Complete step by step solution:
Let us consider that there are three numbers (a-d), a and (a + d) that are in arithmetic progression (A.P).
In the question, we have given that the sum of these three numbers is 33.
Therefore, we can write
\[\;\begin{array}{*{20}{c}}
{ \Rightarrow (a - d) + a + (a + d)}& = &{33}
\end{array}\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow 3a}& = &{33}
\end{array}\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow a}& = &{11}
\end{array}\]
And we have given that the product of these numbers is 792. Therefore, we can write it as,
\[\;\begin{array}{*{20}{c}}
{ \Rightarrow (a - d) \times a \times (a + d)}& = &{792}
\end{array}\]
And
\[\;\begin{array}{*{20}{c}}
{ \Rightarrow a({a^2} - {d^2})}& = &{792}
\end{array}\]
And now put the value of a. Therefore, we will get,
\[\;\begin{array}{*{20}{c}}
{ \Rightarrow 11(121 - {d^2})}& = &{792}
\end{array}\]
\[\;\begin{array}{*{20}{c}}
{ \Rightarrow (121 - {d^2})}& = &{72}
\end{array}\]
\[\;\begin{array}{*{20}{c}}
{ \Rightarrow {d^2}}& = &{121 - 72}
\end{array}\]
\[\;\begin{array}{*{20}{c}}
{ \Rightarrow {d^2}}& = &{49}
\end{array}\]
\[\;\begin{array}{*{20}{c}}
{ \Rightarrow d}& = &7
\end{array}\]
Now, we get the value of a and d. So, we will put these values in the numbers which are,
\[\; \Rightarrow (a - d)\], \[a\]and \[(a + d)\]
Therefore, we will get
(11 – 7), 11, (11 + 7) ……. In A.P
4, 11, 18 …… in A.P
Therefore, the smallest number is 4.
Therefore, the correct option is (A).
Note: Remember that if the numbers are not given in the series which are in arithmetic progression (A.P), then always assume the numbers as (a – d), a, (a + d). In these numbers (a – d) is the smallest number and (a + d) is the greatest number.
Complete step by step solution:
Let us consider that there are three numbers (a-d), a and (a + d) that are in arithmetic progression (A.P).
In the question, we have given that the sum of these three numbers is 33.
Therefore, we can write
\[\;\begin{array}{*{20}{c}}
{ \Rightarrow (a - d) + a + (a + d)}& = &{33}
\end{array}\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow 3a}& = &{33}
\end{array}\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow a}& = &{11}
\end{array}\]
And we have given that the product of these numbers is 792. Therefore, we can write it as,
\[\;\begin{array}{*{20}{c}}
{ \Rightarrow (a - d) \times a \times (a + d)}& = &{792}
\end{array}\]
And
\[\;\begin{array}{*{20}{c}}
{ \Rightarrow a({a^2} - {d^2})}& = &{792}
\end{array}\]
And now put the value of a. Therefore, we will get,
\[\;\begin{array}{*{20}{c}}
{ \Rightarrow 11(121 - {d^2})}& = &{792}
\end{array}\]
\[\;\begin{array}{*{20}{c}}
{ \Rightarrow (121 - {d^2})}& = &{72}
\end{array}\]
\[\;\begin{array}{*{20}{c}}
{ \Rightarrow {d^2}}& = &{121 - 72}
\end{array}\]
\[\;\begin{array}{*{20}{c}}
{ \Rightarrow {d^2}}& = &{49}
\end{array}\]
\[\;\begin{array}{*{20}{c}}
{ \Rightarrow d}& = &7
\end{array}\]
Now, we get the value of a and d. So, we will put these values in the numbers which are,
\[\; \Rightarrow (a - d)\], \[a\]and \[(a + d)\]
Therefore, we will get
(11 – 7), 11, (11 + 7) ……. In A.P
4, 11, 18 …… in A.P
Therefore, the smallest number is 4.
Therefore, the correct option is (A).
Note: Remember that if the numbers are not given in the series which are in arithmetic progression (A.P), then always assume the numbers as (a – d), a, (a + d). In these numbers (a – d) is the smallest number and (a + d) is the greatest number.
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