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Step-by-Step Solution
Step 1: Interpret the Problem
We are asked to insert m Arithmetic Means (A.Ms) and 3 Geometric Means (G.Ms) between 3 and 243. We know that the 4th A.M. is given to be equal to the 2nd G.M. We need to find the value of m.
Step 2: Identify the Formulas for A.M. and G.M.
Arithmetic Progression (A.P.):
If we insert m A.Ms between two numbers a and b, the total number of terms in the resulting A.P. is m + 2. The common difference is given by
d = \frac{b - a}{m + 1} .
Geometric Progression (G.P.):
If we insert n G.Ms between two numbers a and b, the total number of terms in the resulting G.P. is n + 2. The common ratio is given by
r = \left(\frac{b}{a}\right)^{\frac{1}{n + 1}} .
Step 3: Express the 4th A.M.
The first term (a) of the A.P. is 3, and the last term (b) is 243. Using the formula for the common difference:
d = \frac{243 - 3}{m + 1} = \frac{240}{m + 1}.
The 4th A.M. (which is the 4th term in the A.P. sequence) is given by
a + 3d if we count a as the first term, but here we typically say the kth A.M. is a + k·d for k from 1 to m. So the 4th A.M. is:
\text{4th A.M.} = 3 + 4 \cdot \frac{240}{m + 1}.
Step 4: Express the Common Ratio of the G.P.
We are inserting 3 G.Ms between 3 and 243. Thus, the total number of terms in this G.P. is 3 + 2 = 5. The common ratio r is found using:
r = \left(\frac{243}{3}\right)^{\frac{1}{3 + 1}} = \left(\frac{243}{3}\right)^{\frac{1}{4}} = \left(81\right)^{\frac{1}{4}}.
Note that 81 = 3^4 . Hence,
r = \left(3^4\right)^{\frac{1}{4}} = 3.
Step 5: Find the 2nd G.M.
The terms of the geometric sequence starting from 3 with common ratio 3 can be listed as: 3 (first term), 3·3 = 9 (second term), and so on. Hence, the 2nd G.M. after 3 would be:
\text{2nd G.M.} = 3 \times r = 3 \times 3 = 9.
Step 6: Set Up the Equation (4th A.M. = 2nd G.M.)
According to the problem statement,
3 + 4 \times \frac{240}{m + 1} = 9.
Step 7: Solve for m
Simplify the equation step by step:
Subtract 3 from both sides:
4 \times \frac{240}{m + 1} = 9 - 3 = 6.
So,
\frac{960}{m + 1} = 6.
Cross-multiply:
960 = 6 (m + 1).
Divide both sides by 6:
160 = m + 1.
Therefore,
m = 159.
However, based on the originally provided solution, they had:
3 + 4 \times \frac{240}{m + 1} = 3 \cdot 3^2 = 3 \times 9 = 27.
(Not 9.) Let’s carefully verify the 2nd G.M. in the problem statement: The 2nd G.M. was actually the 2nd term in between 3 and 243, which effectively is the 2nd inserted G.M. if we consider 3 as the 1st term and 243 as the 5th term overall.
Hence the G.P. terms look like this for a total of 5 terms:
Term 1: 3
Term 2 (1st G.M inserted): 3·r = 3 × 3 = 9
Term 3 (2nd G.M inserted): 9×3 = 27 (this is the 2nd G.M. we refer to in the problem)
Term 4 (3rd G.M inserted): 27×3 = 81
Term 5: 243
So the 2nd G.M. is actually 27, not 9. We must equate 4th A.M. to 27.
Thus, the equation becomes:
3 + 4 \cdot \frac{240}{m + 1} = 27.
Now simplifying:
Subtract 3 from both sides:
4 \times \frac{240}{m + 1} = 27 - 3 = 24.
So,
\frac{960}{m + 1} = 24.
Cancelling 24:
m + 1 = \frac{960}{24} = 40.
Hence,
m = 39.
Final Answer
m = 39
