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| Question 235791:  Hi! My question is: Find the partial sum S_n for a geometric series such that a_{4} = 216, a_{9} = 52488, and n = 10.   I think I know the formula to use (Sn = a1 (1-r^n) / (1-r)) but 1.) I'm not completely sure and @. I don't know what to plug in where. Thank you very much!
 
 Answer by Theo(13342)
      (Show Source): 
You can put this solution on YOUR website! Your basic formula is correct as shown here !!!!! 
 In a geometric series, each succeeding term is equal to the last term multiplied by a common ratio.
 
 In your geometric series, you are given that:
 
 a[4] = 216
 a[9] = 52488
 
 You need to find a[1] and r.
 
 In general,
 
 a[n] = a[1] * r^(n-1)
 
 and:
 
 a[n] = r*a[n-1]
 
 This second equation can be generalized to
 
 a[n] = r^(n-x) * a[x]
 
 where:
 
 n is the nth term in the series.
 x is the xth term in the series.
 
 In your problem, you have:
 
 a[9] = 52488
 a[4] = 216
 
 This means that, using the generalized formula, we can solve for r as follows:
 
 n = 9
 x = 4
 n-x = 9-4 = 5
 a[9] = 52488
 a[4] = 216
 
 The generalized formula becomes:
 
 52488 = r^5 * 216
 
 Divide both sides of this equation by 216 to get:
 
 52488/216 = r^5
 
 Simplify to get:
 
 243 = r^5
 
 Take the 5th root of both sides to get:
 
 r = 3
 
 Now you want to solve for a[1] which is the first term in your sequence.
 
 We have:
 
 a[1] = ?????
 a[4] = 216
 n = 4
 x = 1
 n-x = 3
 
 Using the generalized formula again, we get:
 
 216 = a[1] * r^3
 
 Solve for a[1] to get:
 
 a[1] = 216 / 3^3 = 8
 
 Your first term in this geometric series is:
 
 a[1] = 8
 
 Your ratio is:
 
 r = 3
 
 The number of terms in your geometric series is:
 
 n = 10
 
 The sum of a geometric series is given by the equation:
 
 S[n] = a[1] * (1-r^n)/(1-r)
 
 The sum of this geometric series becomes:
 
 S[10] = 8 * ((1-3^10)/(1-3)
 
 because:
 
 a[1] = 8
 r = 3
 n = 10
 
 This makes:
 
 S[10] = 8 * ((-59048)/(-2))
 
 Which becomes:
 
 S[10] = 236192
 
 That's your answer.
 
 Some additional information regarding geometric sequences can be found here !!!!! and here!!!!!
 
 FYI,
 
 Since a[1] = 8, we can find a[4] and a[9] using the formula for the nth term in a geometric sequence.
 
 That formula is:
 
 a[n] = a[1] * r^(n-1)
 
 a[4] = 8 * 3^(3) = 8 * 27 = 216
 a[9] = 8 * 3^(8) = 8 * 6561 = 52488
 
 
 
 
 
 
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