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555-605 Level|   Sequences|      
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Bunuel
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G is the sequence of numbers g1, g2, g3…gn such that each term 24 Jul 2020, 04:06
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\(G\) is the sequence of numbers \(g_1, g_2, g_3…g_n\) such that each term following the first is one more than two times the preceding term. If \(g_2 + g_4 = 30 \frac{1}{2}\), what is the first term in the sequence?


A. \(\frac{3}{4}\)

B. \(1\)

C. \(\frac{3}{2}\)

D. \(\frac{9}{4}\)

E. \(\frac{7}{2}\)
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D
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G is the sequence of numbers g1, g2, g3…gn such that each term 24 Jul 2020, 04:24
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each term following the first is one more than two times the preceding term.
Let G1 = x
G2 = 2x+1

G3 = 2G2+1 = 2(2x+1)+1 = 4x+2+1 = 4x+3

G4 = 2G3+1 = 2(4x+3)+1 = 8x+6+1 = 8x+7

G1+G4 = 30½
X+8x+7 = 30.5
9x = 30.5-7
9x = 23.5
G1 = 2.61

Bunuel none of the option is satisfying the constraint. Can we take approximate in exam?

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G is the sequence of numbers g1, g2, g3…gn such that each term 24 Jul 2020, 04:28
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yashikaaggarwal
each term following the first is one more than two times the preceding term.
Let G1 = x
G2 = 2x+1

G3 = 2G2+1 = 2(2x+1)+1 = 4x+2+1 = 4x+3

G4 = 2G3+1 = 2(4x+3)+1 = 8x+6+1 = 8x+7

G1+G4 = 30½
X+8x+7 = 30.5
9x = 30.5-7
9x = 23.5
G1 = 2.61

Bunuel none of the option is satisfying the constraint. Can we take approximate in exam?

Posted from my mobile device

Fixed the typo. It should have been: \(g_2 + g_4 = 30 \frac{1}{2}\)
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G is the sequence of numbers g1, g2, g3…gn such that each term Updated on: 17 Jul 2021, 09:26
Originally posted by BrushMyQuant on 24 Jul 2020, 04:41.
Last edited by BrushMyQuant on 17 Jul 2021, 09:26, edited 1 time in total.
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We are given a sequence \(g_1, g_2, g_3…g_n\) such that "each term following the first is one more than two times the preceding term"

This means that \(g_2\) = 2*\(g_1\) + 1 ...(1)
(one more than two times the previous term)

\(g_3\) = 2*\(g_2\) + 1 = 2(2*\(g_1\) + 1 ) + 1 = 4*\(g_1\) + 3
\(g_4\) = 2*\(g_3\) + 1 = 2*(4*\(g_1\) + 3) + 1 = 8\(g_1\) + 7
\(g_4\) = 8\(g_1\) + 7 ...(2)
And we know that \(g_2 + g_4= 30 \frac{1}{2}\) ...(3)
Substituting value of \(g_2\) and \(g_4\) from (1) and (2) respectively in (3) we get
2*\(g_1\) + 1 + 8\(g_1\) + 7 = 30 \(\frac{1}{2}\)
=> 10*\(g_1\) + 8 = \(\frac{61}{2}\)
=> 10*\(g_1\) = \(\frac{61}{2}\) - 8 = \(\frac{61- 2*8 }{2}\) = \(\frac{45}{2}\)
=> \(g_1\) = \(\frac{45}{2*10}\) = \(\frac{9}{4}\)

So, Answer will be D
Hope it helps!

Watch the following video to learn How to Sequence problems

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G is the sequence of numbers g1, g2, g3…gn such that each term 24 Jul 2020, 04:46
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Bunuel
\(G\) is the sequence of numbers \(g_1, g_2, g_3…g_n\) such that each term following the first is one more than two times the preceding term. If \(g_2 + g_4 = 30 \frac{1}{2}\), what is the first term in the sequence?


A. \(\frac{3}{4}\)

B. \(1\)

C. \(\frac{3}{2}\)

D. \(\frac{9}{4}\)

E. \(\frac{7}{2}\)

First term = g1=X
Given= each term following the first is one more than two times the preceding term

Hence
G2= 2X+1
G3=2G2+1= 4X+3
G4=2G3+1=8X+7

Given G2+G4=61/2
2X+1+8x+7=61/2
By solving we get

G1=X=9/4
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G is the sequence of numbers g1, g2, g3…gn such that each term 24 Jul 2020, 13:35
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Given: \(A_n\) = 2 * \(A_{(n-1)}\) + 1, for n > 1 ......................... (i)

\(G_2\)+ \(G_4\) = 61/2 .............. (ii)

From (i):
\(G_2\) = 2 * \(G_1\) + 1 ................. (iii)

And,

\(G_4\) = 2 * \(G_3\) + 1 = 2 * [2 * \(G_2\) + 1] + 1 = 4 * [2 * \(G_1\) + 1] + 2 + 1 = 8 * \(G_1\) + 7 .................. (iv)

From (ii), (iii), (iv)
2 * \(G_1\) + 1 + 8 * \(G_1\) + 7 = 61/2
10 * \(G_1\) + 8 = 61/2
20 * \(G_1\) + 16 = 61
20 * \(G_1\) = 45
\(G_1\) = 9/4

Answer: D
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G is the sequence of numbers g1, g2, g3…gn such that each term 10 Apr 2023, 03:24
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