In this page the first term of each sequence is denoted a or a1 without any special reason when either of both is used.
Problem 1: Find the first Term
A radio station considered giving away $4000 every day in the month of August for a total of $124,000. Instead, they decided to increase the amount given away every day while still giving away the same total amount. If they want to increase the amount by $100 each day, how much should they give away the first day?
You know the values of n, Sn, and d. Use the sum formula that contains d.
Multiply each side by 2/31.
Subtract 3000 from each side.
Divide each side by 2.
The radio station should give away $2500 the first day.
P2. In an arithmetic series: given ℓ=28, S=144, and there are total 9 terms. Find a.
Here, ℓ=28, S=144 and n=9. The sum of n terms of an arithmetic series is given by
P3. A man saved $33000 in 10 months. In each month after the first, he saved $100 more than he did in the preceding month. How much did he save in the first month?
Let the money saved by the man in the first month be $a.
It is given that in each month afier the first, he saved $100 more than he did in the preceding month. So, the money saved by the man every month is in arithmetic series with common difference $100.
Number of months, n=10.
Sum of money saved in 10 months, S10=$33,000.
Using the formula, Sn =½n[2a+(n-1)d], we get
Hence, the money saved by the man in the first month is $2,850.
P4. In an arithmetic series: given an =4, d=2, Sn =-14, find n and a.
Here, an =4, d=2 and Sn =-14.
The sum of n terms of an arithmetic series is given by
Putting the value of a from equation (1), we get
n+2=0→n=-2, not taken due to (+)n.
Putting the value of d in equation (1), we get a=6-2(7)=-8
P5. A man arranges to pay off debt of 336000 by 40 monthly instalments which form an arithmetic series. When 30 of the installments are paid, he dies leaving one-third of the debt unpaid. Find the value of the first instalment.
Let the value of the first installment be $a.
Since the monthly installments form an arithmetic series, so let us suppose the man increases the value of each installment by $d every month.
∴ Common difference of the arithmetic series =$d.
Amount paid in 30 installments =336,000-⅓⋅$36,000=$36,000-$12,000=$24,000
Let Sn denote the total amount of money paid in the n installments. Then,
2a+39d=1800 … (2)
Subtracting (1) from (2), we get
Putting d=20 in (1), we get
Thus, the value of the first installment is $510.