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Step-by-Step Solution
Step 1: Understand the Given Information
1. We have two geometric progressions (G.P.s):
• \{a_k\} with first term a_1 = 4 and common ratio r_1 .
• \{b_k\} with first term b_1 = 4 and common ratio r_2 .
2. We are also given r_1 < r_2 .
3. Define c_k = a_k + b_k .
4. It is given that c_2 = 5 and c_3 = \tfrac{13}{4} .
5. We want to find the value of \displaystyle \sum_{k=1}^{\infty} c_k \;-\;\bigl(12\,a_6 \;+\;8\,b_4\bigr) .
Step 2: Express Conditions Using the Terms of the G.P.s
• Since a_k = 4 \,(r_1)^{k-1} and b_k = 4 \,(r_2)^{k-1} , we have:
c_k = a_k + b_k = 4\,(r_1)^{k-1} + 4\,(r_2)^{k-1}.
• For k=2 :
c_2 = a_2 + b_2 = 4\,r_1 + 4\,r_2 = 5.
Hence,
r_1 + r_2 = \frac{5}{4}.
• For k=3 :
c_3 = a_3 + b_3 = 4\,r_1^2 + 4\,r_2^2 = \frac{13}{4}.
Hence,
r_1^2 + r_2^2 = \frac{13}{16}.
Step 3: Solve for the Common Ratios r_1 and r_2
1. From r_1 + r_2 = \tfrac{5}{4} , we square both sides:
(r_1 + r_2)^2 = \left(\frac{5}{4}\right)^2 = \frac{25}{16}.
But
(r_1 + r_2)^2 = r_1^2 + 2\,r_1\,r_2 + r_2^2.
2. We also know r_1^2 + r_2^2 = \tfrac{13}{16} , so substituting:
\frac{25}{16} = \frac{13}{16} + 2\,r_1\,r_2
\quad \Longrightarrow \quad
2\,r_1\,r_2 = \frac{25}{16} - \frac{13}{16} = \frac{12}{16} = \frac{3}{4}.
Hence,
r_1\,r_2 = \frac{3}{8}.
3. We now have a system:
r_1 + r_2 = \frac{5}{4},
\quad
r_1\,r_2 = \frac{3}{8}.
Solving this system under the condition r_1 < r_2 gives
r_1 = \frac{1}{2}, \quad r_2 = \frac{3}{4}.
Step 4: Compute the Required Terms a_6 and b_4
• a_6 = 4 \,(r_1)^{5} = 4 \times \left(\frac{1}{2}\right)^5 = 4 \times \frac{1}{32} = \frac{1}{8}.
• b_4 = 4 \,(r_2)^{3} = 4 \times \left(\frac{3}{4}\right)^3 = 4 \times \frac{27}{64} = \frac{27}{16}.
Step 5: Calculate the Infinite Sum of c_k
Since c_k = 4(r_1)^{k-1} + 4(r_2)^{k-1} , the infinite sum is:
\sum_{k=1}^{\infty} c_k
= \sum_{k=1}^{\infty} \bigl[4(r_1)^{k-1}\bigr]
+ \sum_{k=1}^{\infty} \bigl[4(r_2)^{k-1}\bigr].
1. The sum of the G.P. 4(r_1)^{k-1} for k=1 to \infty is
4 \times \frac{1}{1 - r_1}
\quad
\text{since } |r_1| = \tfrac{1}{2} < 1.
Substituting r_1 = \tfrac{1}{2} :
4 \times \frac{1}{1 - \tfrac{1}{2}} = 4 \times 2 = 8.
2. The sum of the G.P. 4(r_2)^{k-1} for k=1 to \infty is
4 \times \frac{1}{1 - r_2}
\quad
\text{since } |r_2| = \tfrac{3}{4} < 1.
Substituting r_2 = \tfrac{3}{4} :
4 \times \frac{1}{1 - \tfrac{3}{4}} = 4 \times 4 = 16.
3. Hence,
\sum_{k=1}^{\infty} c_k = 8 + 16 = 24.
Step 6: Evaluate 12\,a_6 + 8\,b_4
1. From Step 4, a_6 = \tfrac{1}{8} and b_4 = \tfrac{27}{16} .
2. Thus,
12\,a_6 + 8\,b_4
= 12 \times \frac{1}{8}
+ 8 \times \frac{27}{16}
= \frac{12}{8} + \frac{216}{16}.
Simplifying each term:
\frac{12}{8} = 1.5,
\quad
\frac{216}{16} = 13.5.
So,
12\,a_6 + 8\,b_4 = 1.5 + 13.5 = 15.
Step 7: Final Value
Finally, we want
\sum_{k=1}^{\infty} c_k
\;-\;
\bigl(12\,a_6 + 8\,b_4\bigr)
= 24 \;-\; 15
= 9.
Therefore, the required result is \boxed{9}.
