FIND THE NTH TERM FROM GIVEN TWO TERMS OF ARITHMETIC PROGRESSION

Find the n^th term of the arithmetic sequence.

Problem 1 :

a₁ = -4 and a₅ = 16

Solution :

a₁ = -4 and a₅ = 16

Here a₁ and a₅ are 1^st term and 5^th term respectively.

a_n = a₁ + (n – 1)d

a₅ = a₁ + (5 – 1)d

16 = -4 + 4d

Adding 4 on both sides.

16 + 4 = 4d

20 = 4d

Dividing 4 on both sides.

20/4 = 4/4d

5 = d

a_n = a₁ + (n – 1)d

a_n = -4 + (n – 1)5

a_n = -4 + 5n – 5

a_n = 5n – 9

So, the nth term of the sequence is a_n = 5n – 9.

Problem 2 :

a₃ = 94 and a₆ = 85

Solution :

a₃ = 94 and a₆ = 85

a_n = a₁ + (n – 1)d

a₃ = a₁ + (3 – 1)d

94 = a₁ + 2d --- (1)

a₆ = a₁ + (6 – 1)d

85 = a₁ + 5d --- (2)

From (1) and (2)

9 = -3d

Dividing -3 on both sides.

9/-3 = -3/-3d

-3 = d

Substitute (-3) in equation (1)

94 = a₁ + 2(-3)

94 = a₁ – 6

Adding 6 on both sides.

94 + 6 = a₁ – 6 + 6

100 = a₁

a_n = 100 + (n – 1)(-3)

a_n = 100 – 3n + 3

a_n = 97 – 3n

So, the nth term of the sequence is a_n = 97 – 3n.

Problem 3 :

a₅ = 190 and a₁₀ = 115

Solution :

a₅ = 190 and a₁₀ = 115

a_n = a₁ + (n – 1)d

a₅ = a₁ + (5 – 1)d

190 = a₁ + 4d --- (1)

a₁₀ = a₁ + (10 – 1)d

115 = a₁ + 9d --- (2)

From (1) and (2)

75 = -5d

Dividing -5 on both sides.

75/-5 = -5/-5d

-15 = d

Substitute (-15) in equation (1)

190 = a₁ + 4(-15)

190 = a₁ – 60

Adding 60 on both sides.

190 + 60 = a₁ – 60 + 60

250 = a₁

a_n = 250 + (n – 1)(-15)

a_n = 250 – 15n + 15

a_n = 265 – 15n

So, the nth term of the sequence is a_n = 265 – 15n.

Problem 4 :

a₆ = -38 and a₁₁ = -73

Solution :

a₆ = -38 and a₁₁ = -73

a_n = a₁ + (n – 1)d

a₆ = a₁ + (6 – 1)d

-38 = a₁ + 5d --- (1)

a₁₁ = a₁ + (11 – 1)d

-73 = a₁ + 10d --- (2)

From (1) and (2)

35 = -5d

Dividing -5 on both sides.

35/-5 = -5d/-5

-7 = d

Substitute (-7) in equation (1)

-38 = a₁ + 5(-7)

-38 = a₁ – 35

Adding 35 on both sides.

-38 + 35 = a₁ – 35 + 35

-3 = a₁

a_n = -3 + (n – 1)(-7)

a_n = -3 – 7n + 7

a_n = 4 – 7n

 

So, the nth term of the sequence is a_n = 4 – 7n.

Problem 5 :

a₃ = 19 and a₁₅ = -1.7

Solution :

a₃ = 19 and a₁₅ = -1.7

a_n = a₁ + (n – 1)d

a₃ = a₁ + (3 – 1)d

19 = a₁ + 2d --- (1)

a₁₅ = a₁ + (15 – 1)d

-1.7 = a₁ + 14d --- (2)

From (1) and (2)

20.7 = -12d

Dividing -12 on both sides.

20.7/-12 = -12d/-12

-1.725 = d

Substitute (-1.725) in equation (1)

19 = a₁ + 2(-1.725)

19 = a₁ – 3.45

Adding 3.45 on both sides.

19 + 3.45 = a₁ – 3.45 + 3.45

22.45 = a₁

a_n = 22.45 + (n – 1)(-1.725)

a_n = 22.45 – 1.725n + 1.725

a_n = 24.175 – 1.725n

So, the nth term of the sequence is a_n = 24.175 – 1.725n.

Problem 6 :

a₅ = 16 and a₁₄ = 38.5

Solution :

a₅ = 16 and a₁₄ = 38.5

a_n = a₁ + (n – 1)d

a₅ = a₁ + (5 – 1)d

16 = a₁ + 4d --- (1)

a₁₄ = a₁ + (14 – 1)d

38.5 = a₁ + 13d --- (2)

From (1) and (2)

-22.5 = -9d

Dividing -9 on both sides.

-22.5/-9 = -9d/-9

2.5 = d

Substitute (2.5) in equation (1)

16 = a₁ + 4(2.5)

16 = a₁ + 10

Subtracting 10 on both sides.

16 - 10 = a₁ + 10 - 10

6 = a₁

a_n = 6 + (n – 1)(2.5)

a_n = 6 + 2.5n – 2.5

a_n = 3.5 + 2.5n

So, the nth term of the sequence is a_n = 3.5 + 2.5n.