Arithmetic Progression Example.

Q:If the sum of the first 7 terms of an AP is 49 and that of the first 17 terms is 289,find the sum of its first n terms.

Solution:

We know that the Sum of the first n terms of an AP is given by:

Sn = (n / 2) * ((2 * a + (n-1) * d)).

Therefore, as per the first requirement,

49 = (7 / 2) * ((2 * a + (7-1) * d)).

- 98 = 14a + 42d
- 7 = a + 3 d (1)

Similarly, as per the second requirement,

289 = (17 / 2) * ((2 * a + (17-1) * d)).

- 578 = 34a + 272d
- 17 = a + 8d (2)

Now, from (1), we have , a = 7 – 3d. Substituting this value of a in (2), we have

17 = 7-3d + 8d

Whence, 10 = 5d,

Thus, d = 2.

And, then a is 7 – (3 * 2) = 7 – 6 = 1

Thus, we have, a = 1 and d = 2. Now, the sum of the first n terms is

Sn = (n / 2) * ((2 * a + (n-1) * d)).

- Sn = (n / 2) * ((2 * 1 + (n-1) * 2)). ---- Substituting the values of a and d obtained earlier
- Sn = (n / 2) * ((2 + 2n -2))
- Sn = (n / 2) * (2n)
- Sn = n^2

Thus,

The sum of its first n terms = n^2.