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 first q terms of an A.P. is 63q – 3q2.
Also it is given that its pth term is –60,
Now we have to find the value of p. Also, the 11th term of this A.P.
CBSE DHANPAT RAI publications
Class:- 10th
Solutions of CBSE Mathematics
Question 43
Given Sn = 63q – 3q2,
On putting n = 1, we get the first term(a), S1 = a = 63(1) – 3(1)2 = 60.
On putting n = 2 gives S2 = a + a + d = 63(2) – 3(2)2 = 114
=> d = 114 – 2a
=> d = 114 – 120 = –6
The pth term of the A.P., ap = a + (p – 1)d = –60
=> 60 + (p – 1)(–6) = –60
=> (p – 1)(–6) = –120
=> p – 1 = 20
=> p = 21
11th term of the A.P., a11 = a+10d
= 60 + 10(–6)
= 0
Hence, the value of p is 21 and 11th term of the A.P. is 0.