In an AP the sum of the first 7 terms is 49 and the sum of the first 17 terms is 289. Find an expression for the sum of its first n terms.

Class 10Arithmetic ProgressionsAlgebra — Sequences

Step-by-Step Solution

\frac{7}{2} [2 a + 6 d] = 49

gives

a + 3 d = 7

, and

\frac{17}{2} [2 a + 16 d] = 289

gives

a + 8 d = 17

. Subtracting,

5 d = 10

→

d = 2

and

a = 1

. So

S_{n} = \frac{n}{2} [2 + 2 (n - 1)] = \frac{n}{2} (2 n) = n^{2}

— this AP is just the odd numbers 1, 3, 5, ….

Answer: A)

n^{2}

**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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