The 3rd term of an arithmetic progression (AP) is 14 and its 7th term is 34. If the n-th t… — JAMB UTME

The 3rd term of an arithmetic progression (AP) is $14$ and its 7th term is $34$ . If the $n$ -th term of this AP is equal to the 4th term of a geometric progression (GP) whose first term is $2$ and common ratio is $3$ , what is the value of $n$ ?

A
10
11Answer
C
12
D
9

Answer

The value of

n

11

The AP has first term

a = 4

and common difference

d = 5

, giving

T_n = 4 + 5(n-1) = 5n - 1

. The 4th term of the GP is

2 \times 3^{3} = 54

. Setting

5n - 1 = 54

gives

5n = 55

, so

n = 11

Step-by-Step Solution

Find the first term

a

and common difference

d

of the arithmetic progression.

a = 4

and

d = 5

The

n

-th term of an AP is given by

T_n = a + (n-1)d

. Using given terms:

T_3 = a + 2d = 14

and

T_7 = a + 6d = 34

. Subtracting the first equation from the second yields

4d = 20 \Rightarrow d = 5

. Substituting

d = 5

into

a + 2(5) = 14

gives

a = 4

Calculate the 4th term of the geometric progression (

G_4

G_4 = 54

The

m

-th term of a GP is given by

G_m = a_{gp} \cdot r^{m-1}

. With first term

a_{gp} = 2

and ratio

r = 3

G_4 = 2 \cdot 3^{4-1} = 2 \cdot 3^3 = 2 \cdot 27 = 54

Equate

T_n

G_4

and solve for

n

n = 11

Set

T_n = G_4 \Rightarrow 4 + (n-1)5 = 54

. Simplifying gives

(n-1)5 = 50 \Rightarrow n-1 = 10 \Rightarrow n = 11

Key Concept

Solving simultaneous AP/GP equations using the

n

-th term formulas

T_n = a + (n-1)d

and

G_n = a r^{n-1}

Estimated Time:2m 0s

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Key Concept

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