Question

Let \(a_1, a_2, a_3, \dots\) be a G.P. of increasing positive terms such that \(a_2 \cdot a_3 \cdot a_4=64\) and \(a_1 + a_3 + a_5 = \frac{813}{7}\). Then \(a_3 + a_5 + a_7\) is equal to:

• A. 3256
• B. 3248
• C. 3244
• D. 3252

Hint:

In G.P. problems, look for relationships between terms. The product of terms equidistant from the center is constant. For \(a_2, a_3, a_4\), \(a_2 a_4 = a_3^2\), so their product is \(a_3^3\).
Also, notice that the required sum \(a_3 + a_5 + a_7\) is just the given sum \(a_1 + a_3 + a_5\) multiplied by \(r^2\).
This pattern recognition can save calculation steps.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are dealing with a Geometric Progression (G.P.) with first term \(a\) and common ratio \(r\). The terms are positive and increasing, which means \(a>0\) and \(r>1\). We are given two conditions involving the terms of the G.P. and asked to find the value of another sum of terms.
Step 2: Key Formula or Approach:
The \(n\)-th term of a G.P. is given by \(a_n = ar^{n-1}\).
A key property of G.P.s is that for any three consecutive terms \(a_{k-1}, a_k, a_{k+1}\), we have \(a_k^2 = a_{k-1}a_{k+1}\). More generally, \(a_k a_j = a_p a_q\) if \(k+j=p+q\).
Step 3: Detailed Explanation:
Let the first term be \(a_1 = a\) and the common ratio be \(r\).
From the first condition: \(a_2 \cdot a_3 \cdot a_4 = 64\).
Using the formula for G.P. terms: \((ar)(ar^2)(ar^3) = 64\).
\[ a^3 r^{1+2+3} = a^3 r^6 = 64 \] \[ (ar^2)^3 = 4^3 \] \[ ar^2 = 4 \] This means the third term is \(a_3 = 4\).
From the second condition: \(a_1 + a_3 + a_5 = \frac{813}{7}\). \[ a + ar^2 + ar^4 = \frac{813}{7} \] We already know \(a_3 = ar^2 = 4\). Let's substitute this:
\[ a + 4 + (ar^2)r^2 = \frac{813}{7} \] \[ a + 4 + 4r^2 = \frac{813}{7} \] From \(ar^2=4\), we have \(a = \frac{4}{r^2}\). Substitute this into the equation:
\[ \frac{4}{r^2} + 4 + 4r^2 = \frac{813}{7} \] \[ 4\left(\frac{1}{r^2} + 1 + r^2\right) = \frac{813}{7} \] \[ r^2 + \frac{1}{r^2} + 1 = \frac{813}{28} \] \[ r^2 + \frac{1}{r^2} = \frac{813}{28} - 1 = \frac{813 - 28}{28} = \frac{785}{28} \] Let \(y = r^2\). Then \(y + \frac{1}{y} = \frac{785}{28}\).
\[ \frac{y^2+1}{y} = \frac{785}{28} \implies 28y^2 - 785y + 28 = 0 \] Solving the quadratic equation for \(y\): \[ y = \frac{785 \pm \sqrt{(-785)^2 - 4(28)(28)}}{2(28)} = \frac{785 \pm \sqrt{616225 - 3136}}{56} = \frac{785 \pm \sqrt{613089}}{56} \] The square root is \(\sqrt{613089} = 783\). \[ y = \frac{785 \pm 783}{56} \] Two possible values for \(y\): \(y = \frac{1568}{56} = 28\) or \(y = \frac{2}{56} = \frac{1}{28}\).
So, \(r^2 = 28\) or \(r^2 = 1/28\). Since the G.P. is of increasing positive terms, we must have \(r>1\), which implies \(r^2>1\). Therefore, we choose \(r^2 = 28\).
Finally, we need to calculate \(a_3 + a_5 + a_7\):
\[ a_3 + a_5 + a_7 = ar^2 + ar^4 + ar^6 \] We can factor out \(r^2\): \[ = r^2 (a + ar^2 + ar^4) \] We know \(a + ar^2 + ar^4 = a_1+a_3+a_5 = \frac{813}{7}\) and \(r^2 = 28\). \[ a_3 + a_5 + a_7 = 28 \times \frac{813}{7} = 4 \times 813 = 3252 \] Alternatively, we can calculate each term: \[ a_3 = 4 \] \[ a_5 = a_3 r^2 = 4 \times 28 = 112 \] \[ a_7 = a_5 r^2 = 112 \times 28 = 3136 \] \[ a_3 + a_5 + a_7 = 4 + 112 + 3136 = 3252 \] Step 4: Final Answer:
The value of \(a_3 + a_5 + a_7\) is 3252.