9.2 Arithmetic & Geometric Sequences

Key Takeaways

  • Arithmetic and Geometric Sequences is a named item in the NTS Algebra block and appears in both pure-sequence and word-problem form.
  • An arithmetic sequence adds a constant difference: the nth term is a plus (n minus 1) times d.
  • A geometric sequence multiplies by a constant ratio: the nth term is a times r raised to the power (n minus 1).
  • The arithmetic sum is n over 2 times the sum of the first and last terms, which is faster than the expanded formula whenever the last term is known.
  • An infinite geometric series converges to a over (1 minus r) only when the absolute value of r is less than 1.
Last updated: August 2026

Arithmetic & Geometric Sequences

Arithmetic and Geometric Sequences is one of the eight named topics in the NTS Algebra block. Sequence items reward pure formula recall more than almost anything else on the quantitative paper: four formulas cover the entire topic, and once you have them, most items resolve in under a minute.


Telling the Two Apart

Given a list of numbers, test both operations on consecutive terms:

TestIf constantSequence type
Subtract each term from the nextcommon difference $d$Arithmetic (AP)
Divide each term by the previouscommon ratio $r$Geometric (GP)
  • $4,;11,;18,;25,;\ldots$ — differences are all 7 → arithmetic.
  • $3,;12,;48,;192,;\ldots$ — ratios are all 4 → geometric.
  • $2,;6,;24,;120,;\ldots$ — neither; these are factorial-style products $\times 3, \times 4, \times 5$. Some NTS items are deliberately neither, and the answer is found by pattern description rather than by formula.

Arithmetic Sequences

an=a+(n1)dSn=n2[2a+(n1)d]=n2(a+an)a_n = a + (n-1)d \qquad\qquad S_n = \frac{n}{2}\bigl[2a + (n-1)d\bigr] = \frac{n}{2}(a + a_n)

The second form of the sum is the one to reach for whenever you already know the last term.

Worked Example — Finding a Term

The 5th term of an AP is 23 and the 12th term is 58. Find the 20th term.

a+4d=23a+11d=58a + 4d = 23 \qquad a + 11d = 58

Subtracting: $7d = 35$, so $d = 5$ and $a = 23 - 20 = 3$.

a20=3+19(5)=98a_{20} = 3 + 19(5) = 98

Worked Example — Summing

Sum all multiples of 6 between 20 and 200.

First multiple is 24, last is 198, difference 6.

n=198246+1=1746+1=30S30=302(24+198)=15×222=3,330n = \frac{198 - 24}{6} + 1 = \frac{174}{6} + 1 = 30 \qquad S_{30} = \frac{30}{2}(24 + 198) = 15 \times 222 = 3{,}330

The plus one in the count formula is the most-forgotten step in the whole topic.

Arithmetic Mean

The AM of $x$ and $y$ is $\dfrac{x+y}{2}$. To insert $k$ arithmetic means between $x$ and $y$, treat the result as an AP with $k+2$ terms, so $d = \dfrac{y-x}{k+1}$.


Geometric Sequences

an=arn1Sn=a(rn1)r1    (r1)S=a1r    (r<1)a_n = a\,r^{\,n-1} \qquad\qquad S_n = \frac{a(r^{n} - 1)}{r - 1}\;\;(r \neq 1) \qquad\qquad S_\infty = \frac{a}{1 - r}\;\;\bigl(|r| < 1\bigr)

Worked Example — Finding a Term

The 3rd term of a GP is 20 and the 6th term is 160. Find the first term.

ar5ar2=16020=8r3=8r=2\frac{ar^{5}}{ar^{2}} = \frac{160}{20} = 8 \quad\Rightarrow\quad r^{3} = 8 \quad\Rightarrow\quad r = 2 ar2=204a=20a=5ar^{2} = 20 \quad\Rightarrow\quad 4a = 20 \quad\Rightarrow\quad a = 5

The sequence is $5,;10,;20,;40,;80,;160,\ldots$ ✓

Worked Example — Infinite Series

Sum $18 + 6 + 2 + \dfrac{2}{3} + \cdots$

Here $a = 18$ and $r = \frac{1}{3}$. Since $|r| < 1$ the series converges:

S=18113=1823=27S_\infty = \frac{18}{1 - \frac{1}{3}} = \frac{18}{\frac{2}{3}} = 27

If $|r| \ge 1$ the series diverges and has no finite sum. An option offering a finite value for a series with $r = 2$ is always wrong.

Geometric Mean

The GM of two positive numbers is $\sqrt{xy}$. The GM of 4 and 25 is $\sqrt{100} = 10$, whereas their AM is 14.5 — for any two distinct positive numbers, $AM > GM$ always.


Word Problems That Hide a Sequence

SituationModel
Salary rising by a fixed rupee amount each yearAP
Salary rising by a fixed percentage each yearGP with $r = 1 + \frac{p}{100}$
Simple interest accumulatingAP
Compound interest accumulatingGP
Population growing at a constant percentageGP
Seats increasing by a fixed number per rowAP
A ball rebounding to a fixed fraction of its heightGP

Worked Word Problem

A machine bought for Rs. 800,000 depreciates 25% each year. What is its value after 3 years?

Each year retains 75%, so this is a GP with $r = 0.75$:

V3=800000×(0.75)3=800000×0.421875=Rs. 337,500V_3 = 800000 \times (0.75)^{3} = 800000 \times 0.421875 = \text{Rs. }337{,}500

The trap answer is $800000 - (3 \times 200000) = 200{,}000$, which wrongly treats percentage depreciation as a fixed rupee deduction — arithmetic thinking applied to a geometric process.

Twenty-second routine: compute the second-minus-first difference and the second-divided-by-first ratio. Check both against the next pair. Whichever stays constant tells you which pair of formulas to use, and you have solved the classification step before reading the question.

Test Your Knowledge

The 7th term of an arithmetic sequence is 41 and the 15th term is 89. What is the first term?

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Test Your Knowledge

What is the sum of the infinite geometric series 24 + 8 + 8/3 + ...?

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Test Your Knowledge

A tube well's output falls by 20% of the previous year's output every year. If it delivered 500,000 litres per day this year, what will it deliver in the third year from now?

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Test Your Knowledge

How many multiples of 7 lie between 50 and 300, and what is their sum?

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