JEE Corner — Sequences and Series for JEE Main and Advanced

A.P. Essentials — the Rationalised-Out Half

The rationalised textbook chapter keeps only the G.P., but JEE still tests arithmetic progressions freely — the formulas travel from Class 10:

Key Point (A.P.): For first term $a$ and common difference $d$: $a_n = a + (n-1)d$ and

$$S_n = \frac{n}{2}\left[2a + (n-1)d\right] = \frac{n}{2}(a + l)$$

where $l$ is the last term. The arithmetic mean of $a, b$ is $\frac{a+b}{2}$; inserting $n$ A.M.s between $a$ and $b$ uses $d = \frac{b-a}{n+1}$.

Special series (Aryabhatta's sums — also trimmed from the textbook but JEE-assumed):

$$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}, \qquad \sum k^2 = \frac{n(n+1)(2n+1)}{6}, \qquad \sum k^3 = \left[\frac{n(n+1)}{2}\right]^2$$

[JEE Tip] For series like $2 \cdot 4 + 4 \cdot 6 + 6 \cdot 8 + \cdots$, first write the $n$-th term — $a_n = 2n(2n+2) = 4n^2 + 4n$ — then sum with the special series. Term formula first, sums second, always.

The Infinite G.P.

When $|r| < 1$ the powers $r^n \to 0$, and the sum of the whole infinite series settles to a finite value:

Key Point (Sum to infinity):

$$S_\infty = \frac{a}{1 - r}, \qquad |r| < 1$$

— no finite sum exists when $|r| \ge 1$. The picture: $1 + \frac{1}{2} + \frac{1}{4} + \cdots = 2$, the halves filling a length-2 bar.

Recurring decimals are infinite G.P.s: $0.\overline{3} = \frac{3}{10} + \frac{3}{100} + \cdots = \frac{3/10}{1 - 1/10} = \frac{1}{3}$, and $0.\overline{47} = \frac{47/100}{1 - 1/100} = \frac{47}{99}$.

Choosing terms symmetrically: three G.P. terms as $\frac{a}{r}, a, ar$ (product $a^3$); four terms with given product as $\frac{a}{r^3}, \frac{a}{r}, ar, ar^3$. Three A.P. terms as $a - d, a, a + d$ (sum $3a$).

A.M.-G.M. as an optimiser: for a fixed sum $a + b = s$, the product $ab \le \left(\frac{s}{2}\right)^2$ is maximised when $a = b$ — the equality case of $A \ge G$ turned into a tool.

[JEE Tip] "The sum of an infinite G.P. is $S$ and its first term is $a$" instantly gives $r = 1 - \frac{a}{S}$ — one line, no algebra. Check $|r| < 1$ at the end.

Solved Examples

Example 1: A.P. basics

Find the 10th term and the sum of the first 20 terms of the A.P. $2, 7, 12, \ldots$

Solution:

Step 1 — Identify. $a = 2$, $d = 5$.

Step 2 — Term. $a_{10} = a + 9d = 2 + 45 = 47$.

Step 3 — Sum. $S_{20} = \frac{20}{2}\left[2(2) + 19(5)\right] = 10 \times 99 = 990$.

Takeaway: Two formulas, two substitutions — A.P. basics should cost under a minute.

Example 2: Inserting arithmetic means

Insert three arithmetic means between 3 and 19.

Solution:

Step 1 — Count the steps. Three means make 5 terms, so 4 steps: $d = \frac{19 - 3}{4} = 4$.

Step 2 — Write the means. $7, 11, 15$.

Step 3 — Check. $3, 7, 11, 15, 19$ climbs by 4 throughout ✓.

Takeaway: $n$ means → divide the gap by $n + 1$ — same counting rule as geometric insertion.

Example 3: Infinite G.P. — direct

Evaluate $1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots$

Solution:

Step 1 — Identify. $a = 1$, $r = \frac{1}{3}$, and $|r| < 1$ so the infinite sum exists.

Step 2 — Apply $S_\infty = \frac{a}{1-r}$. $\frac{1}{1 - \frac{1}{3}} = \frac{3}{2}$.

Takeaway: Check $|r| < 1$ before quoting $S_\infty$ — the formula is meaningless otherwise.

Example 4: Infinite G.P. — solve for r

An infinite G.P. has first term 8 and sum 32. Find the common ratio.

Solution:

Step 1 — Set up. $\frac{8}{1 - r} = 32$.

Step 2 — Solve. $1 - r = \frac{8}{32} = \frac{1}{4}$: $r = \frac{3}{4}$.

Step 3 — Validate. $\left|\frac{3}{4}\right| < 1$ ✓ — the G.P. genuinely converges.

Takeaway: $r = 1 - \frac{a}{S}$ in one line — then confirm the convergence condition.

Example 5: Recurring decimals

Express (i) $0.\overline{3}$ (ii) $0.\overline{47}$ as fractions using infinite G.P.s.

Solution:

Step 1 — (i) Write the G.P. $0.\overline{3} = \frac{3}{10} + \frac{3}{100} + \cdots$: $a = \frac{3}{10}$, $r = \frac{1}{10}$.

Step 2 — Sum. $\frac{3/10}{1 - 1/10} = \frac{3/10}{9/10} = \frac{1}{3}$.

Step 3 — (ii) Two-digit block. $a = \frac{47}{100}$, $r = \frac{1}{100}$: $\frac{47/100}{99/100} = \frac{47}{99}$.

Takeaway: Repeating block over $99\ldots9$ (one 9 per digit) — the G.P. derivation proves the school rule.

Example 6: Special series — direct sums

Evaluate (i) $1 + 2 + \cdots + 100$ (ii) $1^2 + 2^2 + \cdots + 10^2$ (iii) $1^3 + 2^3 + \cdots + 5^3$.

Solution:

Step 1 — (i). $\frac{100 \times 101}{2} = 5050$.

Step 2 — (ii). $\frac{10 \times 11 \times 21}{6} = \frac{2310}{6} = 385$.

Step 3 — (iii). $\left[\frac{5 \times 6}{2}\right]^2 = 15^2 = 225$.

Takeaway: The cube sum is the square of the first-power sum — memorise all three as one family.

Example 7: Term formula then sum

Find the 20th term of the series $2 \times 4 + 4 \times 6 + 6 \times 8 + \cdots$

Solution:

Step 1 — Write the general term. Factors are $2n$ and $2n + 2$: $a_n = 2n(2n+2) = 4n^2 + 4n$.

Step 2 — Substitute. $a_{20} = 40 \times 42 = 1680$.

Takeaway: ALWAYS write $a_n$ first — every later question about the series flows from it.

Example 8: The same series, summed

Find $S_n$ for $2 \times 4 + 4 \times 6 + 6 \times 8 + \cdots$

Solution:

Step 1 — Split via the special series. $S_n = 4\sum k^2 + 4\sum k = \frac{4n(n+1)(2n+1)}{6} + \frac{4n(n+1)}{2}$.

Step 2 — Factor the common part. $= \frac{2n(n+1)}{3}\left[(2n+1) + 3\right] = \frac{2n(n+1)(2n+4)}{3}$.

Step 3 — Simplify and check. $S_n = \frac{4n(n+1)(n+2)}{3}$; at $n = 1$: $\frac{4 \cdot 2 \cdot 3}{3} = 8 = 2 \times 4$ ✓.

Takeaway: Factor out $n(n+1)$ before expanding — the bracket then collapses in one line.

Example 9: A standard product series

Sum $1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots$ to $n$ terms.

Solution:

Step 1 — General term. $a_k = k(k+1) = k^2 + k$.

Step 2 — Apply the special series. $S_n = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$.

Step 3 — Factor. $= \frac{n(n+1)}{6}\left[(2n+1) + 3\right] = \frac{n(n+1)(n+2)}{3}$.

Takeaway: Products of consecutive integers sum to the next "rising factorial" over 3 — a pattern worth remembering.

Example 10: A.P. and G.P. mixed

Three numbers in A.P. sum to 15. Adding $1, 3, 9$ to them in order produces a G.P. Find the numbers.

Solution:

Step 1 — Symmetric A.P. terms. $a - d, a, a + d$: sum $3a = 15$ gives $a = 5$.

Step 2 — G.P. condition. $(5 + 3)^2 = (6 - d)(14 + d)$: $64 = 84 - 8d - d^2$.

Step 3 — Solve. $d^2 + 8d - 20 = 0$ gives $d = 2$ or $d = -10$.

Step 4 — Write both answers. $3, 5, 7$ (check: $4, 8, 16$ is a G.P. ✓) or $15, 5, -5$ (check: $16, 8, 4$ ✓).

Takeaway: The symmetric choice solves one unknown from the sum instantly; the G.P. middle-term property does the rest.

Example 11: Decimal repdigits

Find the sum of $.6 + .66 + .666 + \cdots$ to $n$ terms.

Solution:

Step 1 — Factor the digit. $.6 = \frac{6}{9}(0.9)$, and in general $S_n = \frac{6}{9}\left[(1 - 10^{-1}) + (1 - 10^{-2}) + \cdots + (1 - 10^{-n})\right]$.

Step 2 — Sum the two parts. The 1's give $n$; the powers give $\frac{1}{9}\left(1 - 10^{-n}\right)$.

Step 3 — Assemble. $S_n = \frac{2}{3}\left[n - \frac{1}{9}\left(1 - 10^{-n}\right)\right]$. Check $n = 2$: $\frac{2}{3}(2 - 0.11) = 1.26 = .6 + .66$ ✓.

Takeaway: Decimal repdigits mirror the integer trick — repunits become $1 - 10^{-k}$ instead of $10^k - 1$.

Example 12: Maximum product from a fixed sum

Two positive numbers have sum 20. What is the maximum possible value of their product?

Solution:

Step 1 — Apply A.M.-G.M. $\sqrt{ab} \le \frac{a+b}{2} = 10$.

Step 2 — Square. $ab \le 100$.

Step 3 — Confirm attainment. Equality holds at $a = b = 10$, and $10 \times 10 = 100$ is achieved: maximum $= 100$.

Takeaway: Fixed sum → maximum product at equality; fixed product → minimum sum at equality. One inequality, two optimisations.

Example 13: Powers stay in G.P.

If $a, b, c, d$ are in G.P., prove $(a^n + b^n), (b^n + c^n), (c^n + d^n)$ are in G.P.

Solution:

Step 1 — Parametrise. $b = ar$, $c = ar^2$, $d = ar^3$.

Step 2 — Factor each bracket. $b^n + c^n = a^nr^n(1 + r^n) = r^n(a^n + b^n)$; likewise $c^n + d^n = r^{2n}(a^n + b^n)$.

Step 3 — Read the ratios. Consecutive ratios both equal $r^n$ — a G.P. ∎

Takeaway: Push the parametrisation through and matching factors appear — the ratio $r^n$ was inevitable.

Example 14: An infinite G.P. from a geometric picture

A ball is dropped from 16 m and rebounds to $\frac{3}{4}$ of its previous height each bounce. Find the total distance travelled.

Solution:

Step 1 — Separate the first drop. The initial 16 m happens once; every rebound height is travelled TWICE (up, then down).

Step 2 — Sum the rebounds. Heights $12, 9, \frac{27}{4}, \ldots$: an infinite G.P. with $a = 12$, $r = \frac{3}{4}$: $S_\infty = \frac{12}{1/4} = 48$.

Step 3 — Assemble. Total $= 16 + 2 \times 48 = 112$ m.

Takeaway: The factor 2 on rebounds (up and down) is the marks-losing subtlety — draw the bounces once.

Example 15: Sum of first n odd numbers

Show $1 + 3 + 5 + \cdots + (2n - 1) = n^2$.

Solution:

Step 1 — Recognise the A.P. $a = 1$, $d = 2$, $n$ terms.

Step 2 — Sum. $S_n = \frac{n}{2}\left[2 + (n-1)2\right] = \frac{n}{2}(2n) = n^2$. ∎

Step 3 — Cross-check. $n = 4$: $1 + 3 + 5 + 7 = 16 = 4^2$ ✓.

Takeaway: The odd numbers tile perfect squares — a two-line A.P. proof of a famous identity.