Question

John saves N$ 3500 the first month and every month later decreases it by N$ 75. a) How much will John save in the 11th month? (5) b) How much will he save after two (2) years?

Answer 1

Answer:

(a) $4250,

(b) $5225.

Step-by-step explanation:

Let n represent number of months.

We have been given that John saves $3500 the first month and every month later decreases it by $75.

First of all, we will find formula of John's savings in n months using arithmetic progression, where, 1st term is 3500 and common difference is 75.

$a_n=a_1+(n-1)d$, where,

n = Number of terms in a sequence,

d = Common difference.

$a_n=3500+(n-1)75$

$a_n=3500+75n-75$

$a_n=3425+75n$

(a) To find John's savings in 11 months, we will substitute $n=11$ in above formula.

$a_{11}=3425+75(11)$

$a_{11}=3425+825$

$a_{11}=4250$

Therefore, John would have saved $4250 in the 11th month.

(b) To find John's savings after 2 years, we will substitute $n=24$ in above formula.

$a_{24}=3425+75(24)$

$a_{24}=3425+1800$

$a_{24}=5225$

Therefore, John would have saved $5225 in 2 years.