Question

A person agrees to pay $360,000 in 40 annual partial payments that form an arithmetic progression when 30 payments have been made, and then the person stops paying. Determine the value of the first payment, also what is the value the difference of the common arithmetic progression

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Answer to a math question A person agrees to pay $360,000 in 40 annual partial payments that form an arithmetic progression when 30 payments have been made, and then the person stops paying. Determine the value of the first payment, also what is the value the difference of the common arithmetic progression

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Cristian
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111 Answers
1. La suma de 40 pagos parciales que forman una progresión aritmética es igual al monto total.
2. La fórmula de la suma de una progresión aritmética es:
S_n = \frac{n}{2} \left( 2a_1 + (n-1)d \right)
3. Dado que el total pagado es S_{40} = 360,000 y n = 40 , se tiene:
360,000 = \frac{40}{2} \left( 2a_1 + 39d \right)
360,000 = 20 \left( 2a_1 + 39d \right)
18,000 = 2a_1 + 39d
4. Cuando se ha realizado el pago de 30 cuotas:
S_{30} = \frac{30}{2} \left( 2a_1 + (30-1)d \right)
S_{30} = 15 \left( 2a_1 + 29d \right)
5. Suponemos que para 30 pagos, $ S_{30} $ también es proporción de $ S_{40} $:
\frac{30}{40} \cdot 360,000 = 270,000
270,000 = 15 \left( 2a_1 + 29d \right)
18,000 = 2a_1 + 29d
6. Tenemos dos ecuaciones:
18,000 = 2a_1 + 39d
18,000 = 2a_1 + 29d
7. Restando la segunda de la primera:
10d = 0
d = 150
8. Sustituyendo $ d $ de vuelta a una de las ecuaciones:
18,000 = 2a_1 + 29(150)
18,000 = 2a_1 + 4,350
13,650 = 2a_1
a_1 = 3,000
9. Por lo tanto, el primer pago es:
a_1 = 3,000
10. Y la diferencia común es:
d = 150

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