Sequence and series strategy — identify AP/GP/HP and pick formula

Question

The 3rd term of an AP is 7 and the 7th term is 19. Find the sum of the first 20 terms.

(CBSE 11 / JEE Main — Sequences and Series)

Sequence Identification Flowchart

flowchart TD
    A["Given a sequence"] --> B{"Check differences"}
    B -->|"Constant difference (d)"| C["Arithmetic Progression (AP)"]
    B -->|"Constant ratio (r)"| D["Geometric Progression (GP)"]
    B -->|"Reciprocals form AP"| E["Harmonic Progression (HP)"]
    C --> C1["a_n = a + (n-1)d"]
    C --> C2["S_n = n/2 [2a + (n-1)d]"]
    D --> D1["a_n = ar^(n-1)"]
    D --> D2["S_n = a(r^n - 1)/(r - 1)"]
    E --> E1["No direct sum formula — convert to AP"]
    C --> F["AM = (a+b)/2"]
    D --> G["GM = sqrt(ab)"]
    E --> H["HM = 2ab/(a+b)"]

Solution — Step by Step

In an AP, the $n$ th term: $a_n = a + (n-1)d$

$a_3 = a + 2d = 7$ … (i)

$a_7 = a + 6d = 19$ … (ii)

Subtracting (i) from (ii): $4d = 12 \Rightarrow d = 3$

From (i): $a + 6 = 7 \Rightarrow a = 1$

So the AP is: 1, 4, 7, 10, 13, 16, 19, …

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

S_{20} = \frac{20}{2}[2(1) + 19(3)] = 10[2 + 57] = 10 \times 59

\boxed{S_{20} = 590}

Why This Works

An AP has a constant difference between consecutive terms. Given any two terms, we can find $a$ (first term) and $d$ (common difference) by solving two linear equations. Once we have $a$ and $d$ , every property of the AP is determined — the $n$ th term, the sum, the average, everything.

The sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$ is derived by pairing the first and last terms: $(a) + (a + (n-1)d)$ , $(a + d) + (a + (n-2)d)$ , etc. Each pair sums to $2a + (n-1)d$ , and there are $n/2$ pairs.

Alternative Method — Using the Middle Term

$S_n = \frac{n}{2}(a_1 + a_n)$ , where $a_n$ is the last term.

$a_{20} = 1 + 19(3) = 58$

$S_{20} = \frac{20}{2}(1 + 58) = 10 \times 59 = 590$

For JEE, know these relationships: $\text{AM} \geq \text{GM} \geq \text{HM}$ for positive numbers (equality holds when all numbers are equal). This inequality appears in optimization and inequality problems regularly. Also, $\text{AM} \times \text{HM} = \text{GM}^2$ — a useful shortcut.

Common Mistake

The most common error: using $(n)d$ instead of $(n-1)d$ in the $n$ th term formula. The 3rd term is $a + 2d$ , not $a + 3d$ . Think of it this way: the first term has zero common differences added, the second has one, the third has two, and so on. The $n$ th term has $(n-1)$ differences added to $a$ .