Arithmetic Sequences

The simplest type of numerical sequence is an arithmetic sequence.

Definition: Arithmetic Sequence
An arithmetic (or linear) sequence is a sequence of numbers in which each new term is calculated by adding a constant value to the previous term

For example, 1, 2, 3, 4, 5, 6, … is an arithmetic sequence because you add 1 to the current term to get the next term:

first term: 1
second term: 2=1+1
third term: 3=2+1
⋮
nth term: n=(n-1)+1

General Equation for the nth-term of an Arithmetic Sequence
More formally, the number we start out with is called a₁ (the first term), and the difference between each successive term is denoted d, called the common difference.

The general arithmetic sequence looks like:

a₁=a₁
a₂=a₁+d
a₃=a₂+d=(a₁+d)+d=a₁+2d
a₄=a₃+d=(a₁+2d)+d=a₁+3d
⋮
a_n=a₁+d⋅(n-1)
Thus, the equation for the nth-term will be:
a_n=a₁+d⋅(n-1)
Given a₁ and the common difference, d, the entire set of numbers belonging to an arithmetic sequence can be generated.

Definition: Arithmetic Sequence
An arithmetic (or linear) sequence is a sequence of numbers in which each new term is calculated by adding a constant value to the previous term:

a_n=a_(n-1)+d
where
● a_n represents the new term, the nth-term, that is calculated;
● a_(n-1)represents the previous term, the (n-1)th-term;
● d represents some constant.

Arithmetic Progression (A.P.)

An arithmetic progression is a sequence of terms such that the difference between any term and the one immediately preceding it is a constant. This difference is called the common difference.

For example, the sequences
(i) 3, 7, 11, 15, 19, …
(ii) 7, 5, 3, 1, -1, …
(iii) 1, ¾, ½, ¼, 0, -¼, …
(iv) 2, 2, 2, 2, …
are arithmetic progressions.
In (i) common difference is 4,
In (ii) common difference is -2,
In (iii) common difference is -¼, and
In (iv) common difference is 0,
In general, an arithmetic progression (A.P.) is given by

a, a+d, a+2d, a+3d, ….
We call a as first term and d as the common difference.
The nth term of the above A.P. is denoted by an and is given by a_n
a_n=a+(n-1)d

nth term or General term (or, last term) of an Arithmetic Progression:
If ‘a‘ be the first term and ‘d‘ be the common difference then

a₁= first term =a=a+(1-1)d
a₂= 2nd term =a+d=a+(2-1)d
a₃= 3rd term =a+2d=a+(3-1))d
a₄= 4th term =a+3d=a+(4-1)d
…
a_n= nth term =a+(n-1)d

a_n=a+(n-1)d

in which a= 1st term, d= common difference, n= number of terms

Example 1: Calculating a Term in an Arithmetic Sequence, Given Two Terms
Two terms in an arithmetic sequence are t₃=4 and t₈=34. What is t₁?
solution:
t₃=4 and t₈=34.
Sketch a diagram. Let the common difference be d.

From the diagram,
t₈=t₃+5d
Substitute: t₈=34, t₃=4
34=4+5d
Solve for d.
30=5d
d=6
Then, t₁=t₃-2d
Substitute: t₃=4, d=6
t₁=4-2(6)
t₁=4-12
t₁=-8

Example 2: Finding a term in an arithmetic sequence
(A) If the first and tenth terms of an arithmetic sequence are 3 and 30, respectively, find the fiftieth term of the sequence.
Solution:
(A) First use Theorem 1 with a₁=3 and a₁₀=30 to find d:

a_n=a₁+(n-1)d
a₁₀=a₁+(10-1)d
30=3+9d
d=3
Now find a₅₀
a₅₀=a₁+(50-1)•3
3+49•3=150

Example 3. Find the 16th and nth terms in an arithmetic sequence with the fourth term 15 and eighth term 37.
Steps:
(1) Write the formula for the nth term of the arithmetic sequence.

t_n=a+(n-1)×d
(2) Substitute n=4 and t₄=15 into the formula.
t₄=a+3d=15
(3) Substitute n=8 and t₈=37 into the formula
t₈=a+7d=37
(4) Solve the simultaneous equations: subtract t₄ from t₈ to eliminate a
(t₈–t₄)…a+7d-a-3d=37-15
4d=22→d=5½
(5) Substitute d=5½ into t₄ and solve for a.
(t₄)…a+3×5½=15
a=-1½
(6) To find the nth term of the arithmetic sequence, substitute the values of a and d into the general formula and simplify.
t_n=-1½+(n-1)×5½
t_n=-1½-5½+5½n
t_n=5½n-7
(7) To find the 16th term, substitute n=16 into the formula established in the previous step and evaluate.
t₁₆=5½×16-7=88-7
t₁₆=81

Example 4. Find the missing terms in this arithmetic sequence: {41, a, 55, b, …}.
Steps:
(1) The first three successive terms are 41, a, 55. Write the rule for the middle term of the three successive terms of an arithmetic sequence.
For x, y, z: y=½(x+z).
(2) Identify the variables.

x=41; y=a; z=55
(3) Substitute the values of x, y and z into the formula in step 1 and evaluate.
a=½(41+55)=48
(4) Find the common difference. (The second term is now known.)
d=t₂–t₁=48-41=7
(5) Find the value of b by adding the common difference to the preceding term.
b=55+7=62
(6) State your answer.
so a=48, b=62

Example 5. The 4th term of an arithmetic sequence is 80 and the 12th term is 32.
a. What is the common difference?
b. What is the first term of the sequence?
solution:
a. How to Proceed
(1) Use a_n=a₁+(n-1)d to write two equation in two variables:

(2) Subtract to eliminate a₁:
(3) Solve for d:
b. Substitute in either equation to find a₁.
80=a₁+3(-6)
80=a₁-18
98=a₁
answer: a. d=-6 b. a₁=98

Example 6. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
Solution:
Here, a₁₁=38 and a₁₆=73 to find a₃₁.
Given that: a₁₁=a+(11-1)d=38

⇒a+10d=38
a=38-10d…(1)
and a₁₆=73 ⇒a+15d=73
Putting the value of a from equation (1), we get 38-10d+15d=73
⇒5d=35
d=7
Putting the value of d in equation (1), we get
a=38-10(7)=-32
Therefore, a₃₁=a+30d=-32+30(7)=178
Hence, the 31st term is 178.

Example 7. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Solution:
Here, a₃=12 and a₅₀=106 to find a₂₉.
Given that: a₃=a+(3-1)d=12

a+2d=12
⇒a=12-2d …(1)
and a₅₀=106
⇒a+49d=106
Putting the value of a from equation (1), we get
12-2d+49d=106
47d=94
d=2
Putting the value of d in equation (1), we get
a=12-2(2)=8
Therefore, a₂₉=a+28d=8+28(2)=64
Hence, the 29th term of the AP is 64.
Let’s read post How to Describe every Arithmetic Sequence?
Ex8. The 2nd term of an arithmetic sequence is 13 and 5th term is 25. What is its 17 term?
(a) 69 (b) 73 (c) 77 (d) 81
Solution:
a₂=a+d=13 … (i)
a₅=a+4d=25 … (ii)
On subtracting (i) from (ii), we get:
3d=12
d=4.
On putting the value of d in (i), we get
a+4=13
a=9.
Now, a₁₇=a+16d=9+16⋅4=73.
Hence, the 17th term is 73.
Answer: (b) 73

Ex9. If 4 times the 4th term of an arithmetic sequence is equal to 18 times its 18th term then find its 22nd term.
Solution:
Let a be the first term and d be the common difference of the AP. Given 4⋅a₄=18⋅a₁₈.

4(a+3d)=18(a+17d) … [a_n=a+(n-1)d]
2(a+3d)=9(a+17d)
2a+6d=9a+153d
7a=-147d
a=-21d
a+21d=0

a₂₂=a+(22-1)d
a₂₂=a+2

1d
a₂₂=0
Hence, the 22nd term of the arithmetic sequence is 0.

Ex10. If 10 times the 10th term of an arithmetic sequence is equal to 15 times the 15th term, show that its 25th term is zero.
Solution:
Let a be the first term and d be the common difference of the AP. Given 10⋅a₁₀=15⋅a₁₅.

10(a+9d)=15(a+14d) …[a_n=a+(n-1)d]
2(a+9d)=3(a+14d)
2a+18d=3a+42d
a=-24d
a+24d=0

a₂₅=a+(25-1)d
a₂₅=a+24d
a₂₅=0
Hence, the 25th term of the arithmetic sequence is 0.

Ex11. The 7th term of an arithmetic sequence is -1 and its 16th term is 17. The nth term of the arithmetic sequence is
(a) (3n+8) (b) (4n-7) (c) (15-2n) (d) (2n-15)
Solution:
Let a be the first term and d be the common difference of the AP. Then,
nth term of the AP, a_n=a+(n-1)d
Given a₇=-1.

a+6d=-1 … (1)
Given a₁₆=17.
a+15d=17 … (2)
Subtracting (1) from (2), we get (a+15d)-(a+6d)=17-(-1)
9d=18
d=2.
Putting d=2 in (1), we get a+6⋅2=-1
a=-1-12=-13
∴ nth term of the AP, a_n=-13+(n-1)⋅2=2n-15
Answer: (d) (2n-15)

Ex12. The 6th and 17th terms of an A.P. are 19 and 41 respectively, find the 40th term. Solution:
Given, a₆=19, a₁₇=41 then

a₆=a+(6-1)d
19=a+5d … (1)
a₁₇=a+(17-1)d
41=a+16d … (2).
Subtract (1) from (2).

Substitute d=2 in (l)
19=a+5⋅2
9=a
∴ 40th term a₄₀=a+(40-1)⋅d
a₄₀=9+39⋅2
=9+78
=87

Ex13. The 10th and 18th terms of an A.P. are 41 and 73 respectively. Find 26th term.
Solution:
Given,

a₁₀=41,a₁₈=73, a_n=a+(n-1)⋅d
⇒a₁₀=a+(10-1)⋅d
41=a+9d …(1)
a₁₈=a+(18-1)d
73=a+17d …(2)
Subtract (1) from (2).

Substitute d=4 in (1).
a+9⋅4=41
a=41-36
a=5
26th term a₂₆=a+(26-1)d
=5+25⋅4
=5+100
=105
∴ 26th term a₂₆=105.