Note:

In this page the first term of each sequence is denotedaora_{1}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*=½

_{n}*n*[2

*a*

_{1}+(

*n*-1)

*d*]

*n*=31,

*d*=100

*S*

_{31}=½⋅31⋅[2

*a*

_{1}+(31-1)100]

*S*

_{31}=124,000

124,000=½⋅31⋅[2

*a*

_{1}+3000]

Multiply each side by 2/31.

*a*

_{1}+3000

Subtract 3000 from each side.

*a*

_{1}

Divide each side by 2.

*a*

_{1}

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*=½

_{n}*n*(

*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, *S*_{10}=$33,000.

Using the formula, *S _{n}* =½

*n*[2

*a*+(

*n*-1)

*d*], we get

*S*

_{10}=½⋅10⋅[2

*a*+(10-1)⋅100]=33000

5(2

*a*+900)=33000

2

*a*+900=6600

2

*a*=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*=-14, find

_{n}*n*and

*a*.

Solution:

Here,

*a*=4,

_{n}*d*=2 and

*S*=-14.

_{n}*a*=

_{n}*a*+(

*n*-1)

*d*

4=

*a*+(

*n*-1)(2)

4=

*a*+2

*n*-2

*a*=6-2

*n*…(1)

The sum of

*n*terms of an arithmetic series is given by

*S*=½

_{n}*n*[2

*a*+(

*n*-1)

*d*]

-14=½

*n*[2

*a*+(

*n*-1)⋅2]

-14=

*n*[

*a*+

*n*-1]

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

*n*[6-2

*n*+

*n*-1]

-14=

*n*[5-

*n*]

-14=5

*n*–

*n*

^{2}

*n*

^{2}-5

*n*-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*=½

_{n}*n*[2

*a*+(

*n*-1)

*d*]

*S*

_{30}=24,000

½⋅30⋅[2

*a*+(30-1)

*d*]=24000

15(2

*a*+29

*d*)=24000

2

*a*+29

*d*=1600… (1)

Also,

*S*

_{40}=$36,000

*a*+(40-1)

*d*]=36000

20(2

*a*+39

*d*)=36000

2

*a*+39

*d*=1800 … (2)

Subtracting (1) from (2), we get

*a*+39

*d*)-(2

*a*+29

*d*)=1800-1600

10

*d*=200

*d*=20.

Putting

*d*=20 in (1), we get

*a*+29⋅20=1600

2

*a*+580=1600

2

*a*=1600-580=1020

*a*=510.

Thus, the value of the first installment is $510.