Two geometric progressions

Insert several numbers between numbers 6 and 384 so that they form with the given GP numbers and that the following applies:
a) the sum of all numbers is 510

And for another GP to apply:

b) the sum of entered numbers is -132
(These are two different geometric sequences but with the same two members)

Correct answer:

n = 4
q = 4
n₂ = 7
q₂ = -2

Step-by-step explanation:

a_{1} = 6 a_{n} = 384 s_{n} = 510 s_{n} = a_{1} \cdot \frac{q ^{n} - 1}{q - 1} a_{n} = a_{1} \cdot q^{n} / q s_{n} = a_{1} \cdot \frac{a _{n} / a _{1} \cdot q - 1}{q - 1} s_{n} \cdot (q - 1) = a_{1} \cdot (a_{n} / a_{1} \cdot q - 1) 510 \cdot (q - 1) = 6 \cdot (384 / 6 \cdot q - 1) 756 q = 3024 q = \frac{3 0 2 4}{7 5 6} = 4 q = 4 n = lo g (a_{n} / a_{1} \cdot q) / lo g q = lo g (384 / 6 \cdot 4) / lo g 4 = 4

s_{2} = - 132 + a_{1} + a_{n} = - 132 + 6 + 384 = 258 s_{2} \cdot (q_{2} - 1) = a_{1} \cdot (a_{n} / a_{1} \cdot q_{2} - 1) 258 \cdot (q_{2} - 1) = 6 \cdot (384 / 6 \cdot q_{2} - 1) 756 q_{2} = - 1512 q_{2} = \frac{- 1 5 1 2}{7 5 6} = - 2 q_{2} = - 2 n_{2} = 7 Verifying Solution: b_{1} = a_{1} = 6 b_{2} = b_{1} \cdot q_{2} = 6 \cdot (- 2) = - 12 b_{3} = b_{2} \cdot q_{2} = (- 12) \cdot (- 2) = 24 b_{4} = b_{3} \cdot q_{2} = 24 \cdot (- 2) = - 48 b_{5} = b_{4} \cdot q_{2} = (- 48) \cdot (- 2) = 96 b_{6} = b_{5} \cdot q_{2} = 96 \cdot (- 2) = - 192 b_{7} = b_{6} \cdot q_{2} = (- 192) \cdot (- 2) = 384 b_{7} = a_{n} S_{2} = b_{2} + b_{3} + b_{4} + b_{5} + b_{6} = (- 12) + 24 + (- 48) + 96 + (- 192) = - 132 n_{2} = lo g (a_{n} / a_{1} \cdot q_{2}) / lo g (q_{2})

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Grade of the word problem:

high school

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