The sixth term of a geometric sequence is 80 more than the fifth term. 1.1 Show that \( a=\frac{80}{r^{3}-r^{4}} \). 2 If it is further given that sum of the fifth and sixth terms is 240 , determine the value of the common ratio.

To solve the problem, we will follow the steps outlined: ### Step 1: Extract Known Conditions 1. The sixth term of a geometric sequence is 80 more than the fifth term. 2. The sum of the fifth and sixth terms is 240. ### Step 1.1: Show that \( a = \frac{80}{r^{3} - r^{4}} \) Let: - \( a \) = the first term of the geometric sequence - \( r \) = the common ratio The \( n \)-th term of a geometric sequence can be expressed as: \[ T_n = a \cdot r^{n-1} \] Thus, we can express the fifth and sixth terms as: - Fifth term: \( T_5 = a \cdot r^{4} \) - Sixth term: \( T_6 = a \cdot r^{5} \) According to the problem, the sixth term is 80 more than the fifth term: \[ T_6 = T_5 + 80 \] Substituting the expressions for \( T_5 \) and \( T_6 \): \[ a \cdot r^{5} = a \cdot r^{4} + 80 \] Now, we can rearrange this equation: \[ a \cdot r^{5} - a \cdot r^{4} = 80 \] Factoring out \( a \): \[ a (r^{5} - r^{4}) = 80 \] This simplifies to: \[ a \cdot r^{4} (r - 1) = 80 \] Now, we can express \( a \): \[ a = \frac{80}{r^{4}(r - 1)} \] Next, we need to express \( a \) in the form \( \frac{80}{r^{3} - r^{4}} \). Notice that: \[ r^{3} - r^{4} = r^{3}(1 - r) \] Thus, we can rewrite \( a \): \[ a = \frac{80}{r^{3}(1 - r)} = \frac{80}{r^{3} - r^{4}} \] ### Step 1.2: Determine the Value of the Common Ratio We are given that the sum of the fifth and sixth terms is 240: \[ T_5 + T_6 = 240 \] Substituting the expressions for \( T_5 \) and \( T_6 \): \[ a \cdot r^{4} + a \cdot r^{5} = 240 \] Factoring out \( a \): \[ a (r^{4} + r^{5}) = 240 \] This can be rewritten as: \[ a r^{4} (1 + r) = 240 \] Now, we have two equations: 1. \( a = \frac{80}{r^{3} - r^{4}} \) 2. \( a r^{4} (1 + r) = 240 \) Substituting the expression for \( a \) from the first equation into the second: \[ \frac{80}{r^{3} - r^{4}} \cdot r^{4} (1 + r) = 240 \] Now, we can solve for \( r \): \[ 80 r^{4} (1 + r) = 240 (r^{3} - r^{4}) \] Dividing both sides by 80: \[ r^{4} (1 + r) = 3 (r^{3} - r^{4}) \] Expanding both sides: \[ r^{4} + r^{5} = 3r^{3} - 3r^{4} \] Combining like terms: \[ r^{5} + 4r^{4} - 3r^{3} = 0 \] Factoring out \( r^{3} \): \[ r^{3} (r + 4 - 3) = 0 \] This gives us: \[ r^{3} (r + 1) = 0 \] Thus, \( r = 0 \) or \( r = -1 \) (not valid for a geometric sequence) or \( r = 3 \). ### Final Answer The common ratio \( r \) is \( 3 \).