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Find the nth term of the sequence 2×3, 4×6, 8×9, 16×12
waec-2018 ssce-exam-2018 waec-mathematics-2018 math 0 Find the nth term of the sequence 2×3, 4×6, 8×9, 16×12

A.  2n×3(n+1)

B.  2n×3n

C.  2n×3n

D.  2n×3n−1

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15 July 2020 University of Lagos Nigeria
12 August 2020 0 Each term has two sequence. That is:

2, 4, 8, 16 -------------- (1)

and,

3, 6, 9, 12 -------------- (2)

Sequence (1) is progressing geometrically and sequence (2) arithmetically.

To find the nth term of a geometric progression, we use the formular: Tn = arn-1,

where a is the first term, r is the common ratio and n is the number of terms.

from sequence (1), a = 2 and r = 4/2 = 2

Thus, the nth term of sequence(1) becomes:

Tn = 2(2)n-1

Tn can also be written as:

Tn = 21 x 2n-1

from indices,

Tn = 21+n-1

Tn = 2n

Hence, the nth term of sequence is Tn = 2n

Similarly, sequence (2) is an arithmetic progression and the formula for the nth term of an arithmetic progression is given as: Tn = a + (n - 1)d

Where a is the first term, d is the common difference and n is the number of terms.

From sequence(2), a = 3 and d = 6 - 3 = 3

Thus the nth term of the arithmetic progression becomes:

Tn = 3 + (n - 1)3

Tn = 3 + 3n - 3

Tn = 3n

Thus, the nth term of sequence (2) is Tn = 3n

Therefore, the nth term of the combined sequence is:

Tn = 2n x 3n

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