CDS Mathematics · Sequence and Series

Progressions and Special Sums

An AP adds a fixed difference, a GP multiplies by a fixed ratio; each has a closed sum, and many harder series telescope.

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Why this matters

Ten PYQs, two of them HARD. Use Sₙ = n/2 (first + last) for an AP and a(rⁿ − 1)/(r − 1) for a GP. The two HARD items are telescoping sums: write the general term as a difference f(n) − f(n + 1) and everything in the middle cancels.

Concept 1 of 2: Sums of APs and GPs

Pair the first and last terms of an AP: every pair has the same sum, so the total is the number of pairs times that sum. A GP's sum follows from multiplying by the ratio and subtracting.

Definition

AP: $a_n = a + (n - 1)d$ ; $S_n = \dfrac n2(a + l) = \dfrac n2[2a + (n - 1)d]$ .
GP: $a_n = ar^{n-1}$ ; $S_n = \dfrac{a(r^n - 1)}{r - 1}$ .
A decreasing AP's sum is largest when you stop at the last non-negative term.
The geometric mean of $3^1, \ldots, 3^7$ is $3$ to the average exponent.
$1 + 2 + \cdots + n = \dfrac{n(n + 1)}{2}$ ; $\sum n^2 = \dfrac{n(n + 1)(2n + 1)}{6}$ ; $\sum n^3 = \left(\dfrac{n(n + 1)}{2}\right)^2$ .

AP sum

S_n = \dfrac{n}{2}(a + l)

Worked example

Savings start at Rs.

500

a month and rise by Rs.

100

each month. After how many months do they total Rs.

9500

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2021 · CDS (II) 2021 — Elementary Mathematics · Q45Moderate

Example 1 · Sequence and Series · Progressions and Special Sums

A person saves Rs. 1000 more than he did the previous year. If he saves Rs. 2000 in the first year, in how many years will he save Rs. 170000 ?

Stop at the last positive term

36, 33, 30, \ldots

the sum grows until the terms turn negative. Stop at the last term that is not negative; adding the negatives only reduces it.

Drill 6 more on sums of aps and gps

Concept 2 of 2: Telescoping sums

If every term is a difference of consecutive values of one function, adding the terms cancels everything in the middle. Only the first and the last survive.

Definition

$\dfrac{1}{n(n + 1)} = \dfrac1n - \dfrac1{n + 1}$ , so $\sum_{1}^{n} = 1 - \dfrac{1}{n + 1}$ .
$\dfrac{2n + 1}{n^2(n + 1)^2} = \dfrac1{n^2} - \dfrac1{(n + 1)^2}$ .
$\sqrt{1 + \dfrac1{n^2} + \dfrac1{(n + 1)^2}} = 1 + \dfrac1n - \dfrac1{n + 1}$ .
For an infinite sum, the last surviving piece tends to $0$ .

Telescoping

\sum_{r=1}^{n} \left(\dfrac1r - \dfrac1{r + 1}\right) = 1 - \dfrac{1}{n + 1}

Worked example

Find

\dfrac{1}{2 \times 3} + \dfrac{1}{3 \times 4} + \cdots + \dfrac{1}{9 \times 10}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2021 · CDS (I) 2021 — Elementary Mathematics · Q30Moderate

Example 2 · Sequence and Series · Progressions and Special Sums

\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \cdots + \frac{1}{n(n + 1)} = \frac{99}{100}

then what is the value of

n

Count the surviving ends

Starting at

\dfrac{1}{2 \times 3}

instead of

\dfrac{1}{1 \times 2}

, the first survivor is

\dfrac12

, not

1

. Write the first two and last two terms out before cancelling.

Drill 2 more on telescoping sums

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Sums of APs and GPs
AP sum
$S_n = \dfrac{n}{2}(a + l)$
Telescoping sums
Telescoping
$\sum_{r=1}^{n} \left(\dfrac1r - \dfrac1{r + 1}\right) = 1 - \dfrac{1}{n + 1}$

Watch out for (2)

Stop at the last positive term
→ Sums of APs and GPs
Count the surviving ends
→ Telescoping sums

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