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 10th term of an A.P. is 21
and the sum of its first 10 terms is 120,
Now we have to find its nth term
CBSE DHANPAT RAI publications
Class:- 10th
Solutions of CBSE Mathematics
Question 29
By using the formula of nth term of an A.P.
an = a + (n – 1)d
So,
10th term of the given A.P., a10 = 21
=> a + 9d = 21 …..(1)
By using the formula of the sum of n terms of an A.P.
Sn = n[2a + (n − 1)d] / 2.
So,
S10 = 10[2a + (10 − 1)d] / 2
=> 120 = 5(2a + 9d)
=> 2a + 9d = 24 …. (2)
On subtracting (1) from (2), we get
=> 2a + 9d – a – 9d = 24 – 21
=> a = 3
On putting a = 3 in eq(1), we get,
=> 3 + 9d = 21
=> 9d = 18
=> d = 2
So, an = 3 + (n – 1)2
= 3 + 2n – 2
= 2n + 1
Hence, the nth term of the given A.P is 2n + 1.