Note:
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?
Solution:
You know the values of n, S_n, and d. Use the sum formula that contains d.

S_n =½n[2a₁+(n-1)d]
n=31,d=100
S₃₁=½⋅31⋅[2a₁+(31-1)100]
S₃₁=124,000
124,000=½⋅31⋅[2a₁+3000]
Multiply each side by 2/31.
8000=2a₁+3000
Subtract 3000 from each side.
5000=2a₁
Divide each side by 2.
2500=a₁
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.
Solution:
Here, ℓ=28, S=144 and n=9. The sum of n terms of an arithmetic series is given by

S_n =½n(a+ℓ)
144=9/2 (a+28)
144×2/9=a+28
32=a+28
a=4

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?
Solution:
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.
∴ d=$100.
Number of months, n=10.
Sum of money saved in 10 months, S₁₀=$33,000.
Using the formula, S_n =½n[2a+(n-1)d], we get

S₁₀=½⋅10⋅[2a+(10-1)⋅100]=33000
5(2a+900)=33000
2a+900=6600
2a=6600-900=5700
a=2850.
Hence, the money saved by the man in the first month is $2,850.

P4. In an arithmetic series: given a_n =4, d=2, S_n =-14, find n and a.
Solution:
Here, a_n =4, d=2 and S_n =-14.

⇒a_n =a+(n-1)d
4=a+(n-1)(2)
4=a+2n-2
a=6-2n …(1)
The sum of n terms of an arithmetic series is given by
S_n =½n[2a+(n-1)d]
-14=½n[2a+(n-1)⋅2]
-14=n[a+n-1]

Putting the value of a from equation (1), we get

-14=n[6-2n+n-1]
-14=n[5-n]
-14=5n–n²
n²-5n-14=0
(n+2)(n-7)=0
n+2=0→n=-2, not taken due to (+)n.
n-7=0→n=7
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.
Solution:
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 S_n denote the total amount of money paid in the n installments. Then,

S_n =½n[2a+(n-1)d]
S₃₀=24,000
½⋅30⋅[2a+(30-1)d]=24000
15(2a+29d)=24000
2a+29d=1600… (1)
Also, S₄₀=$36,000
½⋅40⋅[2a+(40-1)d]=36000
20(2a+39d)=36000
2a+39d=1800 … (2)
Subtracting (1) from (2), we get
(2a+39d)-(2a+29d)=1800-1600
10d=200
d=20.
Putting d=20 in (1), we get
2a+29⋅20=1600
2a+580=1600
2a=1600-580=1020
a=510.
Thus, the value of the first installment is $510.