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Find the n^{th} term of the sequence 2×3, 4×6, 8×9, 16×12

A. 2^{n}×3(n+1)

B. 2^{n}×3n

C. 2^{n}×3^{n }

D. 2^{n}×3^{n−1}

**Comments**

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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: T_{n} = ar^{n-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:

T_{n} = 2(2)^{n-1}

T_{n} can also be written as:

T_{n} = 2^{1} x 2^{n-1}

from indices,

T_{n} = 2^{1+n-1}

T_{n} = 2^{n}

**Hence, the nth term of sequence is T _{n} = 2^{n}**

Similarly, sequence (2) is an arithmetic progression and the formula for the nth term of an arithmetic progression is given as: T_{n} = 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:

T_{n} = 3 + (n - 1)3

T_{n} = 3 + 3n - 3

T_{n} = 3n

**Thus, the nth term of sequence (2) is T _{n} = 3n**

Therefore, the nth term of the combined sequence is:

T_{n} = 2^{n} x 3n

**Ans C**

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