Question

Mike started saving money by putting 50 dollars aside. Each month, he adds more money than the month before. At the end of 36 months, he has saved 4950. How much more does he add each month?

Answer 1

Answer:

He add $5 more each month

Step-by-step explanation:

* Lets consider this problem as an arithmetic sequence because

 he add every month x dollars more

∵ He start with 50 dollars ⇒ a (1st amount)

∵ He add x dollars every month ⇒ d

∵ He did that for 36 months ⇒ n

∵ He saved 4950 dollars  ⇒ Sn

∵ Sn = n/2[2a + (n - 1)d]

* Where Sn is the total money after n months

 a is the first amount he saved

 d is the money he add more each month

∴ 4950 = 36/2[2(50) + (36 - 1)(x)]

∴ 4950 = 18[100 + 35x]

∴ 4950/18 = 100 + 35x

∴ 35x = 275 - 100 = 175

∴ x = 175/35 = 5 dollars

Answer 2

Answer:

Mike added $5 more each month.

Step-by-step explanation:

We are given that Mike started saving money by putting $50 aside. Each month, he adds more money than the previous month and so by the end of 36 months, he saved $4950.

Assuming this to be an arithmetic sequence:

$S_n = \frac{n}{2} (2a+(n-1)d)$

where $S_n=4950$, $n=36$, $a=50$ and $d= x$.

Substituting the given values in the above formula to find how much more money does he add each month.

$4950 = \frac{36}{2} (2 \times 50+(36-1)d)$

$4950=1800+630x$

$x=\frac{3150}{630}$

$x=5$

Therefore, Mike added $5 more each month.