Question

Let a,b be two real numbers between \(3\) and \(81 \)such that the resulting sequence \(3,a,b,81\) is in a geometric progression. The value of \(a+b\) is 

• A. \(36\)
• B. \(29\)
• C. \(90\)
• D. \(27\)
• E. \(81\)

Answer 1

The Correct Option is A

We are given a geometric progression (GP) sequence: \( 3, a, b, 81 \).

Step 1: Understand the GP structure

In a GP, each term is obtained by multiplying the previous term by a common ratio \( r \). Therefore:

\[ a = 3r \]

\[ b = ar = 3r^2 \]

\[ 81 = br = 3r^3 \]

Step 2: Solve for the common ratio \( r \)

From the last equation:

\[ 3r^3 = 81 \]

\[ r^3 = 27 \]

\[ r = \sqrt[3]{27} = 3 \]

Step 3: Find \( a \) and \( b \)

Using \( r = 3 \):

\[ a = 3r = 3 \times 3 = 9 \]

\[ b = 3r^2 = 3 \times 9 = 27 \]

Step 4: Verify the sequence

The sequence becomes \( 3, 9, 27, 81 \), which is indeed a GP with ratio 3.

Step 5: Calculate \( a + b \)

\[ a + b = 9 + 27 = 36 \]

Alternative Approach:

Note that \( r \) could also be negative (\( r = -3 \)):

\[ a = 3 \times (-3) = -9 \]

\[ b = 3 \times (-3)^2 = 27 \]

However, since \( a \) and \( b \) must be between 3 and 81, we discard \( r = -3 \).

Final Answer:

The value of \( a + b \) is \(\boxed{36}\).

Answer 2

The G.P series is:  \(3,a,b,81\) 

means here first term is \(=3\)

last term \(=81\)

So, let 

            \(a=3.r\)

           \(b=3.r=3r^2\)

  Similarly,  \(81=3r^3\)\(\)

              \(⇒ 27=r^3\)\(\)

              \(⇒ r=3\)

Therefore, \(a=3×3=9\)

                \(b=3×3^2=27\)

So,  \(a+b=9+27=36\)