Question: In the arithmetic sequence t1 , t2 , t3 , . . . , tn , . . . , t1 = 23 and tn = tn – 1 – 3 for each n > 1. What is the value of n when tn = – 4 ?

1. -1
2. 7
3. 10
4. 14
5. 20

Correct Answer: C
Solution and Explanation:
Approach Solution 1:

The problem statement states that:

Given:

• In the arithmetic sequence t1 , t2 , t3 , . . . , tn , . . . , t1 = 23 and tn = tn – 1 – 3 for each n > 1.

Find out:

• The value of n when tn = – 4.

In the given sequence, the first term is given.
We can use it to find the second term.
Once the second term is known, we can use that value to find the third term, and so on.

t1 = 23
t2 = t1 – 3 = 23 – 3 = 20
t3 = t2 – 3 = 20 – 3 = 17
t4 = t3 – 3 = 17 – 3 = 14
t5 = t4 – 3 = 14 – 3 = 11
t6 = t5 – 3 = 11 – 3 = 8
t7 = t6 – 3 = 8 – 3 = 5
t8 = t7 – 3 = 5 – 3 = 2
t9 = t8 – 3 = 2 – 3 = -1
t10 = t9 – 3 = -1 – 3 = -4

Therefore, n = 10.
Hence, the value of n when tn is – 4 = 10.

Approach Solution 2:

The problem statement informs that:

Given:

• In the arithmetic sequence t1 , t2 , t3 , . . . , tn , . . . , t1 = 23 and tn = tn – 1 – 3 for each n > 1.

Find out:

• The value of n when tn = – 4.

Note that starting from the second term, each term is 3 less than the previous term. Thus it makes the sequence an arithmetic sequence.
In an arithmetic sequence, the nth term, a_n, can be found by using the formula:
a_n = a_1 + d(n – 1) in which a_1 is the first term and d is the common difference.

Since tn is given, we can modify the formula to:
tn = t1 + d(n – 1) where t1 = 23 and d = -3.

Therefore we get:
tn = t1 + d(n – 1)
-4 = 23 + (-3)(n – 1)
-27 = -3(n – 1)
9 = n – 1
10 = n

Hence, the value of n when tn is – 4 = 10.

Approach Solution 3:

The problem statement implies that:

Given:

• In the arithmetic sequence t1 , t2 , t3 , . . . , tn , . . . , t1 = 23 and tn = tn – 1 – 3 for each n > 1.

Find out:

• The value of n when tn = – 4.

It is a normal question of arithmetic Progression, whose 1st term is 23 and the common difference is -3

Therefore, as per we know:
Tn = a+ (n-1)d
-4 = 23 +(n-1)(-3)
n = 10

Hence, the value of n when tn is – 4 = 10

Approach Solution 4:

The problem statement suggests that:

Given:

• In the arithmetic sequence t1 , t2 , t3 , . . . , tn , . . . , t1 = 23 and tn = tn – 1 – 3 for each n > 1.

Find out:

• The value of n when tn = – 4.

As stated in the question, we know that:
t1=23
Therefore, t2 = t1-3 = 20
Then, t3 = t2-3 =17
and so on...

Here is when we must consider the formula for AP since we know the common difference is -3
tn = t1 + d(n-1)

Given, tn=-4
-4 = 23 + (-3) (n-10)
=> n=10

Hence, the value of n when tn is – 4 = 1

“In the arithmetic sequence t1 , t2 , t3 , . . . , tn , . . . , t1 = 23 and tn”- is a topic of the GMAT Quantitative reasoning section of the GMAT exam. This book has been taken from the book “The Official Guide for GMAT Review”. To solve the GMAT Problem Solving questions, the candidates must have basic knowledge of calculations and mathematics. The candidates can follow GMAT Quant practice papers to practice varieties of questions that will enable them to improve their mathematical skills.

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