If the first four terms of an arithmetic progression are p, p + 2 q, 3

Question

If the first four terms of an arithmetic progression are p, p + 2 q, 3p + q and 30 respectively, find the value of the 2016th term of the progression.

a) 12344
b) 14532
c) 15130
d) 16126
e) 16120

Shrinkhla21 · Answer

The answer is D. 
Using the equation a+(n-1)d= nth term in AP. 
Using this, we get multiple equations and get a and d values. Which can be used to find the 2016th term.

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RaghavKhanna · Answer

a1 = p
a2 = p+2q
a3 = 3p+q
a4 = 30

a4 - a3 = a3 - a2 = a2 - a1

now, a4 - a3,
=> 30 - 3p - q = 2p - q
=> p = 6

a3 - a2 = a2 - a1
=> 2p - q = 2q
q = 2/3p
=> q = 4

Series: 6,14,22,30....
a2016 = a + 2015d = 6 + 2015(8) = 16126

IanStewart · Answer

You don't need to know what an "arithmetic progression" is for the GMAT (if a question includes one, the question will explain that the sequence increases by adding a constant each time). And the GMAT wouldn't ask you to find the "2016th term" in a question like this; that's just complication for the sake of complication, and the GMAT doesn't make questions harder by requiring annoying calculation. A more realistic question might ask for the tenth term, or for some combination of the values of p and q. I'm not sure what the source is, but it's not matching the style of the test.

Here, we know the progression increases by a constant each time. Comparing the first two terms, the sequence is increasing by 2q each time. Since it will increase 2015 times going from the first term to the 2016th term, then adding 2q that many times to p, the term we want to find is p + (2015)(2q) = p + 4030q.

Since the sequence increases by 2q each time, it must increase by 2q going from the third term to the fourth, so from 3p + q to 30, so 3p + 3q = 30, and dividing by 3:

p + q = 10

The sequence must increase by 4q going from the second term to the fourth, so:

p + 6q = 30

Subtracting the first equation from the second leaves us with 5q = 20, or q = 4, from which we find p = 6. So the value we want, p + 4030q, is equal to 6 + 4030(4) = 16126.

luisdicampo · Answer

Deconstructing the Question  
The sequence is an Arithmetic Progression (AP).
Terms:
1.  p 
2.  p + 2q 
3.  3p + q 
4.  30

Target: Find  T_{2016} .

Step 1: Use the Common Difference Property  
In an AP,  T_2 - T_1 = T_3 - T_2 = T_4 - T_3 .

First, compare the differences between the first three terms:
 (p + 2q) - p = (3p + q) - (p + 2q) 
 2q = 2p - q 
 3q = 2p  (Equation 1)

Next, look at the difference leading to the known term (30):
 T_4 - T_3 = T_3 - T_2 
 30 - (3p + q) = (3p + q) - (p + 2q) 
 30 - 3p - q = 2p - q 
Add  q  to both sides:
 30 - 3p = 2p 
 5p = 30 
 p = 6 .

Substitute  p=6  back into Equation 1:
 3q = 2(6) \implies 3q = 12 \implies q = 4 .

Step 2: Determine AP Parameters  
First Term ( a ):  p = 6 .
Second Term:  6 + 2(4) = 14 .
Common Difference ( d ):  14 - 6 = 8 .

Step 3: Calculate the 2016th Term  
Formula:  T_n = a + (n - 1)d 
 T_{2016} = 6 + (2016 - 1)8 
 T_{2016} = 6 + 2015 	imes 8 
 T_{2016} = 6 + 16120 
 T_{2016} = 16126 .

Answer:  d

shivani1351 · Answer

As the first four terms of an arithmetic progression are p, p + 2 q, 3p + q and 30 respectively, the common difference:
p+2q-p = 2q

So, the value of the 2016th term of the progression is p+(2015)2q = p+4030q

Now, the common difference can also be calculated as 30-(3p+q) = 30-3p-q
30-3p-q = 2q
30-3p = 3q
Thus, p+q = 10

q = 10-p
p+4030q = p+4030(10-p)
=> p+40300-4030p

Again, the common difference can also be calculated as (3p+q)-(p+2q) = 2p-q
2p-q = 2q
2p = 3q
Therefore, p = 3q/2

3q/2 + q = 10
q = 4
Therefore, p = 6

p+40300-4030p = 6+40300-(4030*6) = 16126

Hence, the correct answer is Option D (Ans)

If the first four terms of an arithmetic progression are p, p + 2 q, 3

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