In the infinite sequence a1, a2, ..., an, an equals the sum of all pre

Question

In the infinite sequence  a_1 ,  a_2 , ...,  a_n ,  a_n  equals the sum of all previous values in the sequence for all values n>2. If  a_1=1  and  a_3=3 , what is the value of  \frac{a_{100}}{a_{97}} ?

A. 3
B. 8
C. 12
D. 27
E. 48

Kurtosis · Accepted Answer

Given:  a_{1} = 1; a_{3} = 3; a_{2} = a_{3} - a_{1} = 2

Sequence: 1, 2, 3, 6, 12, 24 ........ --> Notice that for n > 3, each term is double the previous term
i.e.  a_{6} = 2a_{5} = 4a_{4}

Similarly,  a_{100} = 2a_{99} = 4a_{98} = 8a_{97}

a_{100}/a_{97} = 8{a_{97}/a_{97}} = 8

Answer: B

KalyanVaddagiri · Answer

Using the GP, the CR is 2

General Sum formula for GP =b1*r^n-1/r-1

First term b1 =1; r =2

a100/a97 =2^99/2^96 =2^3 =8 =B

Bang2919 · Answer

I see that the GP sum formula is   / 1 - r. Can you please elaborate on how you got the mentioned formula?

Pratyaksh2791 · Answer

Hi,

The formula which you're talking about is used when R<1
Here, we have R>1 which is 2 so that formula is used

Sent from my vivo 1609 using  GMAT Club Forum mobile app

sriharsha4444 · Answer

we can do this using GP like others mentioned but it will be bit more  tedious .

a1, a2, dont form the GP, so ignore them 
starting from a3, a4,... they all are in gp.

Proof: 
a100 = a1+a2+....a99
we know that a1+a2+...a98 = a99 (a99 comes by summing the previous terms)
so, 
a100 = a99 + a99
Therefore, a100 = 2(a99). Similarly, a99 = 2(a98) and so on until a3.

a100 = a1 + a2 + (a3 + ... a98)
we know that a3 ... a98 are in GP. 
so their sum= a3 * (1-2^95)/(1-2)

numerator will be a1 + a2 + a3 * (1-2^95)/(1-2)

a3 + a3 * (1-2^95)/(1-2)

a3 + (a3 - a3 * 2^95)/ -1

a3 + a3* 2^95 -a3 => a3 * 2^95

We can do the same for denominator and you will get => a3 * 2^92

dividing both => 2^3 => 8

bumpbot · Answer

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In the infinite sequence a1, a2, ..., an, an equals the sum of all pre

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In the infinite sequence a1, a2, ..., an, an equals the sum of all pre

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