0Arithmetic 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.
