Sequences and Series

The chapter opens by defining a sequence as an ordered list of numbers indexed by position, distinguishing finite from infinite sequences and explicit formulas (like $a_n = 2n$) from recursive definitions (like the Fibonacci rule $a_n = a_{n-1} + a_{n-2}$). It then defines the associated series and introduces sigma notation for writing sums compactly. A short recap of the arithmetic progression — general term and sum formula — sets up the central comparison of the chapter: constant addition versus constant multiplication. The bulk of the chapter is then the geometric progression: its general term $a_n = ar^{n-1}$, the sum to $n$ terms $S_n$ (split carefully into the $r=1$ and $r \neq 1$ cases), and, as a natural extension, the sum to infinitely many terms when $|r| < 1$. The geometric mean of two numbers and the technique of inserting several geometric means between two numbers follow directly from the general-term formula. The relationship between A.M. and G.M. — the inequality $A \ge G$ — is proved algebraically and used to solve "given the A.M. and G.M., find the two numbers" problems. The chapter closes with special series: closed-form sums for the first $n$ natural numbers, their squares, and their cubes, each derived by the telescoping-sum technique rather than assumed.

Sequences: definition, notation and the general term

A sequence is an arrangement of numbers in a definite order — formally, a function whose domain is the set of natural numbers $\mathbb{N}$ or a subset of the form $\{1, 2, 3, \ldots, k\}$. The terms are written $a_1, a_2, a_3, \ldots, a_n, \ldots$, where the subscript records the position of the term, not its value; $a_n$, the term at the $n$th position, is called the general term of the sequence, and knowing $a_n$ as a function of $n$ fixes the entire sequence. A sequence with a fixed, finite number of terms is a finite sequence (the ten generations of a person's ancestors, $a_1 = 2, a_2 = 4, \ldots, a_{10} = 1024$, is finite); a sequence that never ends is an infinite sequence (the successive quotients $3, 3.3, 3.33, 3.333, \ldots$ obtained by dividing $10$ by $3$ never terminate).

where: these are explicit formulas — substitute $n = 1, 2, 3, \ldots$ directly to generate every term (NCERT Example 1 builds exactly this skill: for $a_n = 2n+5$, substituting $n=1,2,3$ gives $a_1=7, a_2=9, a_3=11$; for $a_n = \frac{n-3}{4}$, the first three terms are $-\frac12, -\frac14, 0$). Not every sequence has an explicit formula — the sequence of primes $2, 3, 5, 7, 11, \ldots$ has no known algebraic formula for its $n$th term at all and can only be described in words. Many sequences instead obey a recursive rule, where each term is defined in terms of the ones before it: the Fibonacci sequence $1, 1, 2, 3, 5, 8, 13, \ldots$ is generated by $a_1 = a_2 = 1$ and $a_n = a_{n-1} + a_{n-2}$ for $n > 2$ (NCERT Example 3 works a similar recursive case, $a_1 = 1$, $a_n = a_{n-1} + 2$, giving $1, 3, 5, 7, 9$). Both explicit and recursive definitions count as a valid sequence — what matters is that there is some rule, however stated, that determines every term.

Series and sigma notation

Given a sequence $a_1, a_2, a_3, \ldots, a_n$, the expression formed by adding its terms, $a_1 + a_2 + a_3 + \cdots + a_n$, is called the series associated with that sequence. A series is finite or infinite exactly according to whether the underlying sequence is finite or infinite. It is worth being precise about a common source of confusion: the sequence is the ordered list, the series is the indicated sum, and "the sum of the series" refers to the single number you get after actually adding — for the finite series $1 + 3 + 5 + 7$, the sum is $16$.

where: $\Sigma$ (capital sigma) is read "the sum of", $k$ is the running index taking every integer value from the lower limit ($1$) to the upper limit ($n$), and $a_k$ is the general term being summed — sigma notation is nothing more than shorthand for a long addition, and it is the notation used throughout the special-series results at the end of this chapter.

A quick recap: arithmetic progression

Before contrasting it with the geometric progression, recall the arithmetic progression (A.P.) you built in Class 10: a sequence in which every term after the first is obtained by adding a fixed number, the common difference $d$, to the term before it — $a, a+d, a+2d, a+3d, \ldots$. The general term and the sum to $n$ terms are the two working formulas you will need again in this chapter, especially when a problem mixes a G.P. with an A.P. (as several of this chapter's own miscellaneous problems do — for instance, subtracting fixed amounts from three numbers in G.P. to leave three numbers in A.P.).

where: $a$ is the first term, $d = a_{k+1} - a_k$ is the constant common difference (the defining test of an A.P. — the difference between consecutive terms is constant, in contrast with the G.P.'s constant ratio), $n$ is the number of terms, and $l = a_n$ is the last term of a finite A.P. The arithmetic mean (A.M.) of two numbers $a$ and $b$ is the single number $A$ that makes $a, A, b$ an A.P., which forces $A = \frac{a+b}{2}$ — the ordinary average, and the quantity compared against the geometric mean later in this chapter.

Geometric progression: definition and the general term

Now contrast that constant addition with constant multiplication. Consider the sequences $2, 4, 8, 16, \ldots$ and $\frac19, -\frac13, 1, -3, \ldots$ (reversed for illustration) — in each, every term except the first bears a constant ratio to the term immediately before it: $2$ in the first case. Such a sequence is called a geometric progression, abbreviated G.P. Formally, a sequence $a_1, a_2, a_3, \ldots$ (every term non-zero) is a G.P. if

where: $r$ is the common ratio and $a = a_1$ is the first term; writing $a_1 = a$, the G.P. reads $a, ar, ar^2, ar^3, \ldots$. The general-term formula $a_n = ar^{n-1}$ follows the same pattern-spotting NCERT uses to introduce it: $a_2 = ar = ar^{2-1}$, $a_3 = a_2 r = ar^2 = ar^{3-1}$, $a_4 = a_3 r = ar^3 = ar^{4-1}$, and so the exponent is always one less than the term's position (NCERT Example 4 applies this directly: for the G.P. $5, 25, 125, \ldots$, $a = 5$, $r = 5$, so $a_{10} = 5(5)^{9} = 5^{10}$ and, in general, $a_n = 5(5)^{n-1} = 5^n$). A finite G.P. is written $a, ar, ar^2, \ldots, ar^{n-1}$; an infinite one drops the final term and continues with dots.

Sum to n terms of a G.P.

The whole practical value of a G.P. lies in being able to add a large number of its terms without adding them one by one. Denote the sum of the first $n$ terms by $S_n$.

where: $a$ is the first term, $r$ is the common ratio, and $n$ is the number of terms summed. The two cases are genuinely different formulas, not one written two ways — when $r=1$ every term equals $a$, so the sum is simply $n$ copies of $a$ added, and the $r \neq 1$ formula would in fact divide by zero if you tried to force $r=1$ into it. The full derivation of the $r \neq 1$ case, by the classic "multiply by $r$ and subtract" trick, is given below as the first derivation. NCERT Example 7 shows the formula in action for $1 + \frac23 + \frac49 + \cdots$ (here $a=1$, $r=\frac23$): $S_n = 3\left[1 - \left(\frac23\right)^n\right]$, and in particular $S_5 = \frac{211}{81}$.

Sum to infinity of a G.P.

Push the finite-sum formula further: what happens as $n$ grows without bound? If $|r| < 1$, then $r^n \to 0$ as $n \to \infty$ (each multiplication by $r$ shrinks the term further toward zero), so the finite sum $S_n = \frac{a(1-r^n)}{1-r}$ approaches a fixed, finite limit instead of growing forever.

where: this sum exists — the series converges — only when $|r| < 1$; for $|r| \ge 1$ the terms do not shrink to zero and the sum grows without bound or oscillates, so $S_\infty$ is simply not defined. This is exactly the mechanism behind a recurring decimal: $0.\overline{7} = 0.7 + 0.07 + 0.007 + \cdots$ is a G.P. with $a = 0.7$ and $r = 0.1$, so $S_\infty = \frac{0.7}{1-0.1} = \frac{0.7}{0.9} = \frac{7}{9}$ — the familiar fraction for the repeating decimal, produced entirely from the infinite-G.P. formula.

Geometric mean and inserting means

The geometric mean (G.M.) of two positive numbers $a$ and $b$ is defined as $G = \sqrt{ab}$ — the number that makes $a, G, b$ a three-term G.P., since $\frac{G}{a} = \frac{b}{G}$ gives exactly $G^2 = ab$. The idea generalises: given two positive numbers $a$ and $b$, any number of terms can be inserted between them so that the whole extended list is a G.P.

where: with $n$ numbers inserted, $b$ becomes the $(n+2)$th term of the sequence $a, G_1, G_2, \ldots, G_n, b$, so $b = ar^{n+1}$, which is solved for $r$ as shown. NCERT Example 12 works the case $n=3$ between $1$ and $256$: here $b = ar^4 = 256$, giving $r = \pm 4$ (taking real roots), so $r=4$ gives $G_1, G_2, G_3 = 4, 16, 64$, and $r=-4$ gives $-4, 16, -64$ — a reminder that an even-power equation for $r$ has more than one real solution.

The relationship between A.M. and G.M.

Let $A$ and $G$ be the arithmetic mean and geometric mean, respectively, of two given positive real numbers $a$ and $b$: $A = \frac{a+b}{2}$ and $G = \sqrt{ab}$. These two "average-like" numbers are not equal in general, and the precise relationship between them is one of the most quoted inequalities in the chapter.

where: the inequality holds for any two positive reals $a, b$, and it becomes an equality in exactly one situation — when $a$ and $b$ are equal, so that both averages collapse to the same number. The full algebraic proof is the second derivation below; NCERT Example 13 shows the reverse use of the relationship — given $A = 10$ and $G = 8$, recovering the original numbers $a=4, b=16$ (or $a=16,b=4$) by solving $a+b=20$ and $ab=64$ together.

Special series: sums of natural numbers, squares and cubes

Not every series you meet is arithmetic or geometric — a very common type is a series built from powers of the natural numbers themselves, $1, 2, 3, \ldots, n$. Three such sums recur constantly, in this chapter's problems and throughout Class 11–12 mathematics, and each has a clean closed form.

where: the first result is simply the sum of an A.P. with $a=1, d=1, l=n$, applied to $(3)$; the second and third are proved by the telescoping-sum technique — writing a difference of consecutive cubes (or fourth powers) so that a sum collapses, worked in full as the third derivation below. Notice the elegant fact hidden in the third formula: the sum of the first $n$ cubes equals the square of the sum of the first $n$ natural numbers, $\sum k^3 = \left(\sum k\right)^2$.

Board level - the tenth term of a G.P. given two later terms (NCERT Example 6 pattern)

In a G.P., the 3rd term is $24$ and the 6th term is $192$. Find the 10th term.

Step 1. Translate both given terms into the general-term formula $(4)$: $a_3 = ar^2 = 24$, labelled (I), and $a_6 = ar^5 = 192$, labelled (II).

Step 2. Divide (II) by (I) to eliminate $a$ in one step: $\dfrac{ar^5}{ar^2} = \dfrac{192}{24}$, which simplifies to $r^3 = 8$, so $r = 2$.

Step 3. Substitute $r=2$ back into (I) to find $a$: $a(2)^2 = 24 \Rightarrow 4a = 24 \Rightarrow a = 6$.

Step 4. Apply the general-term formula once more for $n=10$: $a_{10} = ar^{9} = 6(2)^9 = 6 \times 512$. Answer: the 10th term is $a_{10} = 3072$.

Board level - recovering two numbers from their A.M. and G.M. (NCERT Example 13 pattern)

The A.M. and G.M. of two positive numbers $a$ and $b$ are $10$ and $8$ respectively. Find the numbers.

Step 1. Write both conditions as equations using the definitions in $(3)$ and the G.M. definition: $\dfrac{a+b}{2} = 10 \Rightarrow a+b = 20$, labelled (I); $\sqrt{ab} = 8 \Rightarrow ab = 64$, labelled (II).

Step 2. Two numbers are known once both their sum and product are known, using the identity $(a-b)^2 = (a+b)^2 - 4ab$. Substitute: $(a-b)^2 = (20)^2 - 4(64) = 400 - 256 = 144$.

Step 3. Take the square root, keeping both signs since $(a-b)^2=144$ has two real roots: $a - b = \pm 12$, labelled (III).

Step 4. Solve (I) and (III) together. Taking $a-b=12$ with $a+b=20$ gives $2a=32 \Rightarrow a=16, b=4$; taking $a-b=-12$ gives $a=4, b=16$ instead — the same pair of numbers in the other order. Answer: the numbers are $4$ and $16$ (in either order), and indeed $\frac{4+16}{2}=10$ and $\sqrt{4 \times 16} = 8$, confirming $A=10 \ge G=8$ as the A.M.–G.M. inequality demands.

Competitive level (JEE / NEET) - a functional equation hiding a geometric progression

A function $f$ satisfies $f(x+y) = f(x)f(y)$ for all $x, y \in \mathbb{N}$, with $f(1) = 3$. If $\sum_{x=1}^{n} f(x) = 120$, find the value of $n$.

Step 1. Decode the functional equation. Setting $y=1$ repeatedly, $f(x+1) = f(x)f(1) = 3f(x)$ — every term is exactly $3$ times the one before it. That is precisely the defining property of a G.P. with common ratio $r = f(1) = 3$, and first term $f(1) = 3$. So $f(1), f(2), f(3), \ldots = 3, 9, 27, \ldots$, i.e. $f(x) = 3^x$.

Step 2. Recognise $\sum_{x=1}^n f(x)$ as the sum of the first $n$ terms of this G.P. and apply formula $(5)$ with $a=3, r=3$: $S_n = \dfrac{3(3^n - 1)}{3-1} = \dfrac{3(3^n-1)}{2}$.

Step 3. Set this equal to the given total and solve for $n$: $\dfrac{3(3^n-1)}{2} = 120 \Rightarrow 3(3^n-1) = 240 \Rightarrow 3^n - 1 = 80 \Rightarrow 3^n = 81$.

Step 4. Since $81 = 3^4$, and the exponential function $3^n$ is one-to-one (strictly increasing) for positive $n$, this equation has exactly one solution. Answer: $n = 4$. Timing note: under two minutes once the functional equation is decoded. The entire problem is a disguise — the moment $f(x+1) = f(1)\cdot f(x)$ is spotted, it collapses into a completely ordinary G.P.-sum question. The trap is trying to solve the functional equation "from scratch" instead of testing the single substitution $y=1$, which is the standard opening move for any equation of the form $f(x+y) = f(x)f(y)$.

The most damaging slip in this chapter is applying the G.P. sum formula $S_n = \frac{a(r^n-1)}{r-1}$ blindly when $r=1$, where it divides by zero and is simply undefined — always check $r$ first and switch to $S_n = na$ if $r=1$. A second very common error is testing the "constant ratio" condition using only the first two terms of a sequence and assuming it holds throughout; the definition requires $a_{k+1}/a_k$ to equal the same $r$ for every consecutive pair, so a sequence like $2, 4, 9, 18, \ldots$ (ratios $2, 2.25, 2$) is not a G.P. even though its first ratio looks promising. Sign errors are frequent with a negative common ratio: if $r<0$, then $r^{n-1}$ alternates sign depending on whether $n$ is odd or even, so $a_n$ oscillates in sign, and forgetting this leads to wrong-signed answers in problems like Example 12's $r=\pm4$ case. Taking an even root (square root, fourth root, and so on) to solve for $r$ or for a mean routinely loses a valid solution — always keep both $\pm$ roots until the problem's context (for example, "positive numbers only") rules one out. Students often confuse the arithmetic mean and geometric mean formulas under exam pressure, writing $\sqrt{a+b}$ or $\frac{ab}{2}$ instead of $\frac{a+b}{2}$ and $\sqrt{ab}$ — keeping the mnemonic "A.M. adds, then halves; G.M. multiplies, then roots" avoids this. Finally, the infinite-sum formula $S_\infty = \frac{a}{1-r}$ is sometimes applied to a G.P. with $|r| \ge 1$, where no finite sum exists at all — always check $|r|<1$ before reaching for it, and recall that finite-G.P. sums have no such restriction on $r$.

• State the definition of a sequence as an ordered list (or function on $\mathbb{N}$), and distinguish an explicit general-term formula from a recursive definition.
• Translate between a series written in full and the same series written in sigma notation, in both directions.
• Recall the A.P. formulas $a_n = a+(n-1)d$ and $S_n = \frac{n}{2}[2a+(n-1)d]$ from Class 10 well enough to use them inside a mixed A.P./G.P. problem.
• Recognise a geometric progression by testing whether $a_{k+1}/a_k$ is constant across every consecutive pair, not just the first two terms.
• Write the general term $a_n = ar^{n-1}$ of a G.P. and solve for any one of $a$, $r$, $n$ or $a_n$ given the other three.
• Derive and correctly apply the sum-to-$n$-terms formula for a G.P., choosing the right case at $r=1$ versus $r \neq 1$.
• State the condition $|r|<1$ for an infinite G.P. to have a sum, and compute $S_\infty = \frac{a}{1-r}$ when it applies.
• Compute the geometric mean $\sqrt{ab}$ of two positive numbers, and insert $n$ geometric means between two given numbers.
• Prove the inequality $A \ge G$ from the single fact that a real square is non-negative, and use it to recover two numbers from their given A.M. and G.M.
• State and apply the three special-series formulas for $\sum k$, $\sum k^2$ and $\sum k^3$, and explain how the telescoping-sum technique produces them.

Ask the tutor: Want to work through NCERT's four-number G.P.-detection problem — where a sum-of-squares inequality forces $a,b,c,d$ into a geometric progression — using the same A.M.–G.M. proof technique step by step?