Three consecutive terms of an arithmetic progression are x + 2, 3x - 1, and 4x + 1. What i… — JAMB UTME

Three consecutive terms of an arithmetic progression are $x + 2$ , $3x - 1$ , and $4x + 1$ . What is the $10^{\text{th}}$ term of the progression?

The

10^{\text{th}}

term of the progression is 70.

Equating the differences between consecutive terms gives

(3x - 1) - (x + 2) = (4x + 1) - (3x - 1)

, which simplifies to

2x - 3 = x + 2

, giving

x = 5

. The first term is

a = 5 + 2 = 7

and the common difference is

d = 14 - 7 = 7

. Substituting these into

T_n = a + (n - 1)d

for

n = 10

yields

T_{10} = 7 + 9(7) = 70

Set up an equation using the common difference property of an arithmetic progression.

(3x - 1) - (x + 2) = (4x + 1) - (3x - 1)

In any arithmetic progression, the difference between consecutive terms is constant (

T_2 - T_1 = T_3 - T_2

Simplify and solve for

x

2x - 3 = x + 2 \implies x = 5

Subtract

x

from both sides and add 3 to both sides.

Find the first term

a

and the common difference

d

First term

a = 5 + 2 = 7

; second term

T_2 = 3(5) - 1 = 14

; common difference

d = 14 - 7 = 7

Substitute

x = 5

into the expressions for the terms.

Calculate the

10^{\text{th}}

term using

T_n = a + (n - 1)d

T_{10} = 7 + (10 - 1)(7) = 7 + 9(7) = 7 + 63 = 70

Apply

n = 10

a = 7

, and

d = 7

to the

n^{\text{th}}

term formula.

Arithmetic Progression - Consecutive terms and nth term evaluation

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