G is the sequence of numbers g1, g2, g3…gn such that each term

Question

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}

yashikaaggarwal · Answer

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

Bunuel · Answer

Fixed the typo. It should have been:  g_2 + g_4 = 30 \frac{1}{2}

BrushMyQuant · Answer

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

https://www.youtube.com/watch?v=EfNT2JoE5CM

Varunsawhney8 · Answer

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

RaghavKhanna · Answer

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 *  G_2  + 1] + 1 = 4 *  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

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