Question

If \( p \) and \( q \) are the real numbers such that the 7th term in the expansion of \( \left( \frac{5}{p^3} \cdot \frac{3q}{7} \right)^8 \) is 700, then \( 49p^2 = \):

• A. \( 4q^2 \)
• B. \( 9q^2 \)
• C. \( 16q^2 \)
• D. \( 25q^2 \) \bigskip

Hint:

When solving for variables in binomial expansions, simplify the relevant term first and then use the known values to solve for the unknown variable.

Answer 1

The Correct Option is B

We are given that the 7th term in the expansion of \( \left( \frac{5}{p^3} \cdot \frac{3q}{7} \right)^8 \) is 700, and we are asked to determine the value of \( 49p^2 \).

Step 1: The general term in the binomial expansion of \( (a + b)^n \) is given by: \[ T_r = \binom{n}{r} a^{n-r} b^r. \] We can rewrite the expression \( \left( \frac{5}{p^3} \cdot \frac{3q}{7} \right)^8 \) as: \[ \left( \frac{5}{p^3} + \frac{3q}{7} \right)^8. \] In this case, \( a = \frac{5}{p^3} \) and \( b = \frac{3q}{7} \). The general term will therefore be: \[ T_r = \binom{8}{r} \left( \frac{5}{p^3} \right)^{8-r} \left( \frac{3q}{7} \right)^r. \] We are interested in the 7th term, corresponding to \( r = 6 \) (since the index starts from 0). 

Step 2: Substitute \( r = 6 \) into the formula for \( T_r \): \[ T_7 = \binom{8}{6} \left( \frac{5}{p^3} \right)^2 \left( \frac{3q}{7} \right)^6. \] Simplifying this gives: \[ T_7 = \binom{8}{6} \left( \frac{5^2}{p^6} \right) \left( \frac{(3q)^6}{7^6} \right). \] Since \( \binom{8}{6} = 28 \), we have: \[ T_7 = 28 \cdot \frac{25}{p^6} \cdot \frac{729q^6}{117649}. \] 

Step 3: Simplify the expression for \( T_7 \): \[ T_7 = 28 \cdot \frac{25 \cdot 729q^6}{p^6 \cdot 117649}. \] We are told that \( T_7 = 700 \), so we set the equation equal to 700: \[ 28 \cdot \frac{25 \cdot 729q^6}{p^6 \cdot 117649} = 700. \] Simplifying this gives: \[ \frac{28 \cdot 25 \cdot 729q^6}{p^6 \cdot 117649} = 700. \] \[ \frac{28 \cdot 25 \cdot 729q^6}{117649} = 700 \cdot p^6. \] \[ \frac{28 \cdot 25 \cdot 729}{117649} = 700 \cdot p^6 \cdot q^{-6}. \] 

Step 4: Rearranging the terms gives a value for \( q^2 \). Solving for \( p \) yields \( 49p^2 = 9q^2 \). Thus, the value of \( 49p^2 \) is \( \boxed{9q^2} \). \bigskip