Lesson: Introduction to Sequences

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Sir i am having doubt in Solving this problem

Each term in sequence S is determined by multiplying the prior term by 2 and dividing that product by 3. What is the 100th term of the sequence S?

(1) The sum of the first 2 terms is 15
(2) The first term of the sequence is 9
gmat-admin's picture

Hi Rajkumar,

I just posted a solution to that question here: https://gmatclub.com/forum/each-term-in-sequence-s-is-determined-by-mult...

Cheers,
Brent

Sir, but in statement 2,
If we multiply the first term by 2/3 we are getting the second term as 6, and next terms 4,8/3......etc.Here the common difference varies.Hence how it is possible to find 100th term?
gmat-admin's picture

Here's how:

term1 = 9
term2 = (9)(2/3) = 6
term3 = (9)(2/3)(2/3) = 4
term4 = (9)(2/3)(2/3)(2/3) = 8/3
term5 = (9)(2/3)(2/3)(2/3)(2/3) = 16/9
.
.
.
If you recognize that we can keep doing this indefinitely, then you can see that we COULD determine the value of term100

We might also see the pattern and recognize that term100 = (9)(2/3)^99

Hi Brent,
A little stuck on this problem: I don’t know how to finish it after the third step below

A, B, C, D, E, F, G, and H are all integers, listed in order of increasing size. When these numbers are arranged on a number line, the distance between any two consecutive numbers is constant. If G and H are equal to 5^12 and 5^13, respectively, what is the value of A?

a) -24(5^12)
b) -23(5^12)
c) -24(5^6)
d) 23(5^12)
5) 24(5^12)

The nth term of an a.p. is given by: a+(n−1)d
a = first term (happen to be looking for A)
n = nth term we are looking for (A is the eighth term)
d = the common difference (4 x 5^12 is the common difference)
a + (8-1)4 x 5^12
a + 7(4 x 5^12)
a + 28 x 5^12
gmat-admin's picture

You're right to say the difference between any two successive values is 4(5^12). However, I think there's a problem with how you set up your equation.

Here's one approach that doesn't rely on our knowledge of arithmetic progressions:

Given: G and H are equal to 5^12 and 5^13 respectively

5^13 - 5^12 = 5^12(5 - 1) = (5^12)(4) = 4(5^12)

So, the DIFFERENCE between any two successive values is 4(5^12).

In other words, to find the NEXT term in the sequence, we just ADD 4(5^12) to the term before for it.

Conversely, to find the PREVIOUS term in the sequence, we just SUBTRACT 4(5^12).

We have:

H = 5^13
G = 5^12
F = 5^12 - 4(5^12)
E = 5^12 - 4(5^12) - 4(5^12) = 5^12 - 8(5^12)
D = 5^12 - 8(5^12) - 4(5^12) = 5^12 - 12(5^12)
Following this PATTERN, we get:
C = 5^12 - 16(5^12)
B = 5^12 - 20(5^12)
A = 5^12 - 24(5^12)

5^12 - 24(5^12) = 1(5^12) - 24(5^12)
= -23(5^12)
Answer: B

Does that help?

Cheers,
Brent

A.....(4)(5^12)..... B.....(4)(5^12)..... C....(4)(5^12)... D......(4)(5^12)..... E.....(4)(5^12)..... F.....(4)(5^12).... G.....(4)(5^12)....H

5^12 – (4)( 5^12) x 6 (times 6 because it’s 6 steps to A from F
5^12 – (24)( 5^12)
= -23(512)
Got it! But I Don’t understand why the equation “a+(n−1)d” won’t work? It’s always worked. Is it because it is going backwards (negative)?

I find this on the Forum:

The user is obviously using the equation I have always used but his setup is a little different as he seems to be equating it to 5^12. The part where he is equating it to 5^12 is where I am a little stuck/confused.


G = A + (n-1) d
5^12 = A + (7-1) 4 (5^12)
As n=7, because its the 7th term in the sequence

Therefore,
A = (5^12) - 24 (5^12)
= (5^12) (1-24) Taking 5^12 common
= - 23 (5^12)
= Answer choice B
gmat-admin's picture

Quote: "I Don’t understand why the equation “a+(n−1)d” won’t work? It’s always worked. Is it because it is going backwards (negative)?"

The formula will work; it just takes some fiddling at the end.

We know that the difference (d) = 4(5^12)

At this point, we can apply the formula term_n = a + (n−1)d, where a = the first term

If we recognize that G (aka 5^12) is the same as term_7, then we can write:

So, term_7 = a + (7−1)[4(5^12)]

Since term_7 = 5^12, we can write: 5^12 = a + (7−1)[4(5^12)]
Simplify: 5^12 = a + (6)[4(5^12)]
Simplify: 5^12 = a + 24(5^12)
Solve for a to get: a = -23(5^12)

----THAT'S THE EASIER VERSION------------

Notice that it's more work if we use term_8 (aka H)

term_8 = 5^13
So, we can write: 5^13 = a + (8−1)[4(5^12)]
Simplify: 5^13 = a + (7)[4(5^12)]
Simplify: 5^13 = a + 28(5^12)

Now what?

From here, we need to recognize that 5^13 = 5(5^12)

So, we can write: 5(5^12) = a + 28(5^12)
So, a = 5(5^12) - 28(5^12) = -23(5^12)

Cheers,
Brent

Never mind that last comment. I was way overthinking it somehow for far too long without realizing/seeing the obvious.The nth term of an a.p. is given by: a+(n−1)d or n = a+(n−1)d. N=N.

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