The sum of the first n terms of an arithmetic progression (A.P.) is given by S_n = 3n^2 + … — JAMB UTME

The sum of the first $n$ terms of an arithmetic progression (A.P.) is given by $S_n = 3n^2 + 5n$ . A geometric progression (G.P.) has a first term of $1$ and a common ratio of $2$ . If the $5^{\text{th}}$ term of the A.P. is equal to the $k^{\text{th}}$ term of the G.P., what is the value of $k$ ?

A
5
6Answer
C
4
D
7

Answer

The value of

k

is 6.

Evaluating the

5^{\text{th}}

term of the A.P. gives

32

. Equating this to the

k^{\text{th}}

term formula of the G.P.,

1 \cdot 2^{k-1} = 32

, leads to

2^{k-1} = 2^5

, which gives

k = 6

Step-by-Step Solution

Find the general formula for the

n^{\text{th}}

term

T_n

of the A.P. using

S_n

T_n = S_n - S_{n-1} = (3n^2 + 5n) - [3(n-1)^2 + 5(n-1)] = 6n + 2

The

n^{\text{th}}

term of a series is the difference between the sum of the first

n

terms and the sum of the first

n-1

terms.

Calculate the

5^{\text{th}}

term of the A.P.

T_5 = 6(5) + 2 = 32

Substitute

n = 5

into the derived expression for

T_n

Set up the equation for the

k^{\text{th}}

term of the G.P.

G_k = a \cdot r^{k-1} = 1 \cdot 2^{k-1} = 2^{k-1}

The standard formula for the

k^{\text{th}}

term of a G.P. is

G_k = a \cdot r^{k-1}

Equate

T_5

and

G_k

to solve for

k

2^{k-1} = 32 \implies 2^{k-1} = 2^5 \implies k - 1 = 5 \implies k = 6

Since the bases are equal (

2

), equate the exponents to find

k

Key Concept

Relating Arithmetic Progression sum formulas to term values and solving indexed Geometric Progression equations.

Estimated Time:2m 0s

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