Today’s topic is Arithmetic Progression (AP). An AP is a sequence of numbers such that the difference between the consecutive terms is constant.

For example:

2, 4, 6, 8…

-1, 10, 21, 32…

4, 3, 2, 1, 0, -1, -2, -3

-1/2, -3/2, -5/2, -7/2…

and so on.

Note that the numbers could be increasing or decreasing. As long as the difference between the consecutive terms is constant, it is an AP. The general form of an AP is given by

a, a+d, a+2d, a+3d…

I hope you see how we get it. ‘a’ is the first term and ‘d’ is the common difference. The nth term has the form ‘a + (n-1)d’. The first term is given by ‘a + (1-1)d’ = ‘a’

The second term is given by ‘a + (2-1)d’ = ‘a + d’ etc

The sum of n terms will be found by doing the following:

a + (a+d) + (a+2d) + … + (a+(n-1)d) = na + (d + 2d + 3d + … + (n-1)d) (summing the n first terms together and clubbing the common differences together in a bracket)

= na + d(1 + 2 + 3 + …(n-1)) (taking d common)

= na + d(n-1)n/2 (this is because the sum of n consecutive integers starting from 1 is given by n(n+1)/2. We will work on this concept in detail next week.)

= n/2 (2a + (n-1)d)

If you notice carefully, 2a + (n-1)d can be re-written as: [a] + [a + (n-1)d] i.e. the sum of first and last terms. So basically the sum of n terms of the AP is [n * (First term + Last term)/2] which is the same as [number of terms * Average of the first and the last terms].

Questions on APs are very simple. Sometimes, people just don’t realize that the question is based on an AP. The questions in GMAT may not give you that the sequence is an AP but it is not tough to figure out. Let us look at an example now.

Question 1: The nth term, t(n), of a certain sequence is defined as t(n) = t(n-1) + 4. If t(1) = -11, then t(82) =

(A) 313

(B) 317

(C) 320

(D) 321

(E) 340

Solution:

The given relation says that every nth term is 4 more than the previous term. So basically, it tells us that the sequence is an AP. Whew! (An AP is very easy to work with)

What is the nth term of an AP? It’s [a + (n-1)d]

What is the 82^{nd} term of this AP? It’s [-11 + 81*4] = 313

Answer (A)

Question 2: If S is the infinite sequence such that t(1) = 4, t(2) = 10, …, t(n) = t(n-1) + 6,…, what is the sum of all the terms from t(10) to t(18)?

(A) 671

(B) 711

(C) 738

(D) 826

(E) 991

Solution: Again, since the nth term is 6 more than the previous term, it is an AP.

We need to find the following sum: t(10) + t(11) + t(12) + … + t(18)

But we only know how to find the sum starting from t(1). Let’s manipulate what we have to find a little to make it similar to what we know.

t(10) + t(11) + t(12) + … + t(18) = Sum of first 18 terms – Sum of first 9 terms

Sum of first 18 terms = (18/2)(2*4 + 17*6)

Sum of first 9 terms = (9/2)(2*4 + 8*6)

t(10) + t(11) + t(12) + … + t(18) = (18/2)(2*4 + 17*6) – (9/2)(2*4 + 8*6)

= 990 – 252 = 738

Answer (C)

As you see, AP questions are easy to work with. Next week, we will discuss some properties of a specific type of AP i.e. consecutive integers. Till then, keep practicing!

*Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the **GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

## Quarter Wit, Quarter Wisdom: Progressing to Arithmetic Progressions

*Posted on March 19, 2012, filed in: GMAT, Quarter Wit, Quarter Wisdom*

Related posts:

- Quarter Wit, Quarter Wisdom: Scrutinizing Sequences
- Quarter Wit, Quarter Wisdom: Scrutinizing a 700+ Level Question on Sequences
- Quarter Wit, Quarter Wisdom: Comparing Roots and Exponents
- Quarter Wit, Quarter Wisdom: A Remainders Post for the Geek in You!
- Quarter Wit, Quarter Wisdom: Exponents - The Bane of Your Preparation II

Thanks :)

Beside AP will there be any questions from GP or HP?

sorry forgot to mention in GMAT.

GMAT may not specifically say that the sequence is an AP/GP but it could give you these sequences. Identifying such sequences and quickly solving for them could help you save time. I am discussing GP as well on the blog though I don’t see much gain in discussing HP.

The nth term, t(n), of a certain sequence is defined as t(n) = t(n-1) + 4. If t(1) = -11, then t(82) =

I don’t quite understand….the expression for an AP is T(n) = a + (n-1)d. Where a is the first term or t(1). In the question stem above, We are given the expression t(n) = 4 + (n-1)t. So I’m assuming that a=4. But it’s also mentioned that t(1) = -11?

Given t(n) = t(n-1) + 4,

at t=1(first term)

t(1) = t(1-1) +4

t(1) = 4

but t is actually the difference between elements in the set and a = -11.

What am I missing here?

Thanks!

Thanks again :)

Hi,

I actually used a slightly different approach,

T(10) = a + 9*6 = 4 + 54= 58

T(18) = 4 + 17*6 =106

so using the basic formula of AP which is n * [ (First term + last term) /2 ]

and considering the sequence of t(10) as the first term we in total 9 terms, with first term as 58 and last term as 106.

Therefore Sum = 9*[ 58 + 106 / 2] =9*164/2 =738.

Hope it helps