Question

Find the four numbers in G.P, whose sum is \[85\] and the product is \[4096\].
A.\[64,16,4,1\]
B.\[64,16,40,1\]
C.\[64,19,4,1\]
D.\[67,16,4,1\]

Answer 1

Hint: Using the concept of G.P as let the first term be a and general ratio be r, then nth term can be given as \[{{\text{T}}_{\text{n}}}{\text{ = a}}{{\text{r}}^{{\text{n - 1}}}}\]. So from the given concept form all the terms and then solve it using the condition given in the question.

Complete step by step answer:

Let the four terms of a G.P be ,
\[{\text{a,ar,a}}{{\text{r}}^{\text{2}}}{\text{,a}}{{\text{r}}^{\text{3}}}\]
Now, as per the given that sum of all the terms is \[85\]
\[
   \Rightarrow {\text{a + ar + a}}{{\text{r}}^{\text{2}}}{\text{ + a}}{{\text{r}}^{\text{3}}} = 85 \\
   \Rightarrow {\text{a(1 + r + }}{{\text{r}}^{\text{2}}}{\text{ + }}{{\text{r}}^{\text{3}}}{\text{) = 85}} \\
 \]
And the product of the four terms is \[4096\]
\[
   \Rightarrow {\text{a(ar)(a}}{{\text{r}}^{\text{2}}}{\text{)(a}}{{\text{r}}^{\text{3}}}) = 4096 \\
   \Rightarrow {\text{(}}{{\text{a}}^4}{{\text{r}}^6}{\text{) = 4096}} \\
 \]
Calculating the factors of
\[
   \Rightarrow {\text{4096 = }}{{\text{a}}^{\text{4}}}{{\text{r}}^{\text{6}}}{\text{ = }}{{\text{2}}^{{\text{12}}}} \\
   \Rightarrow {{\text{a}}^{\text{4}}}{{\text{r}}^{\text{6}}}{\text{ = 1}}{\text{.}}{{\text{4}}^{\text{6}}} \\
 \Rightarrow {\text{a = 1,r = 4}} \\
 \]
And so from this we can conclude that the terms are,
\[
  {\text{a = 1,r = 4}} \\
  {\text{1,4,16,64}} \\
 \]
Hence, option (a) is the correct answer.

Note: In mathematics, a geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.
Properties :
If all the terms of G.P are multiplied or divided by the same non-zero constant then the sequence remains in G.P with the same common ratio.
The reciprocals of the terms of a given G.P. form a G.P