This is the basic and conceptual question from Chapter name- Arithmetic Progression
Chapter number- 9
Exercise 9.6
In this question we have been given that the sum of 7 terms of an A.P. is 49 and that of 17 terms is 289
Now we have to find the sum of n terms.
CBSE DHANPAT RAI publications
Class:- 10th
Solutions of CBSE Mathematics
Question 22
We know sum of n terms of an A.P. is given by Sn = n[2a + (n − 1)d]/2.
So we get, S7 = 49
=> 7[2a + (7 − 1)d]/2 = 49
=> 7[a + 3d] = 49
=> a + 3d = 7 ….. (1)
And also, S17 = 289
=> 17[2a + (17 − 1)d]/2 = 289
=> 17[a + 8d] = 289
=> a + 8d = 17 ….. (2)
On subtracting eq(1) from (2), we get,
=> a + 8d − (a + 3d) = 17 − 7
=> 5d = 10
=> d = 2
On putting d = 2 in (1), we get,
=> a + 3(2) = 7
=> a = 1
Here a = 1, d = 2, so sum of n terms would be,
Sn = n[2(1) + (n − 1)(2)]/2
= n[2 + 2n − 2]/2
= n[n] = n2
Hence, the sum of n terms of the given A.P is n2.