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# In a geometric sequence, each term is a constant multiple of the prece

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In a geometric sequence, each term is a constant multiple of the prece  [#permalink]

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12 Mar 2020, 04:52
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In a geometric sequence, each term is a constant multiple of the preceding one. What is the sum of the first four numbers in a geometric sequence whose second number is 4 and whose third number is 6?

(A) 16

(B) 19

(C) 22$$\frac{1}{2}$$

(D) 21$$\frac{2}{3}$$

(E) 20

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Re: In a geometric sequence, each term is a constant multiple of the prece  [#permalink]

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12 Mar 2020, 04:58
In a geometric sequence, each term is a constant multiple of the preceding one. What is the sum of the first four numbers in a geometric sequence whose second number is 4 and whose third number is 6?

(A) 16

(B) 19

(C) 22$$\frac{1}{2}$$

(D) 21$$\frac{2}{3}$$

(E) 20

GP is written as

a, ar, ar^2, ar^3... and so on

where r = common ratio = Second term/FIrst term = Third term / Second term = ... and so on

$$Here, r = 6/4 = 3/2$$

T_2 = 4 = ar
I.e. $$a= 4/(3/2) = (8/3) = T_1$$

T_3 = 6

and $$T_4 = 6*r = 6*(3/2) = 9$$

Sum, T_1 + T_2 + T_3 + T_4 = (8/3) + 4 + 6 + 9 = 19 8/3 = 21 2/3

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Re: In a geometric sequence, each term is a constant multiple of the prece  [#permalink]

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15 Mar 2020, 03:45
In a geometric sequence, each term is a constant multiple of the preceding one. What is the sum of the first four numbers in a geometric sequence whose second number is 4 and whose third number is 6?

(A) 16

(B) 19

(C) 22$$\frac{1}{2}$$

(D) 21$$\frac{2}{3}$$

(E) 20

Since the second term is 4 and the third is 6, the constant multiple is therefore 6/4 = 3/2.

Going backward from the second term, the first term is 4/(3/2) = 4 * 2/3 = 8/3. Going forward from the third term, the fourth term is 6 x 3/2 = 9. Therefore, the sum of the first four terms is:

8/3 + 4 + 6 + 9 = 2 ⅔ + 19 = 21 ⅔

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Re: In a geometric sequence, each term is a constant multiple of the prece   [#permalink] 15 Mar 2020, 03:45