For the AP 3, 7, 11, 15, …, find the sum of the terms from the 11th to the 20th (both included).

Class 10Arithmetic ProgressionsAlgebra — Sequences

Step-by-Step Solution

a = 3, d = 4

, so

a_{11} = 3 + 40 = 43

and

a_{20} = 3 + 76 = 79

. Those ten terms form an AP of their own, so their sum is

\frac{10}{2} (43 + 79) = 5 \times 122 = 610

. Equivalently

S_{20} - S_{10} = 820 - 210 = 610

Answer: D) 610

**Sₙ =

(n /2) \times

[

2 a + (n - 1) d

]**

Or equivalently, **Sₙ =

(n /2) \times (a + l)

**, where

l

is the LAST term (the n-th term).

**Why these are equal**:

l = a + (n - 1) d

, so

a + l = 2 a + (n - 1) d

.

**Example**: Sum of 1, 4, 7, …, 28.
First find

n : a = 1, d = 3, l = 28

. From aₙ = 28:

1 + 3 (n - 1) = 28

→

n = 10

.
Then S₁₀ =

(10/2) (1 + 28) = 5 \times 29 = 145

.

**Special case

(n

natural numbers)**: Sₙ =

n (n + 1) /2

.
**First

n

odd numbers**: Sₙ =

n^{2}

.
**First

n

even numbers**: Sₙ =

n (n + 1)

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