If the 10th term of an A.P. is 21 and the sum of its first 10 terms i

1.	If the 10th term of an A.P. is 21 and the sum of its first 10 terms is 120, find its nth term.
Answer» Let’s consider a to be the first term and d be the common difference. And we know that, sum of first n terms is: S_n = \(\frac{n}{2}\)(2a + (n − 1)d) and n^th term is given by: a_n = a + (n – 1)d Now, from the question we have S₁₀ = 120 ⟹ 120 = \(\frac{10}{2}\)(2a + (10 − 1)d) ⟹ 120 = 5(2a + 9d) ⟹ 24 = 2a + 9d …. (1) Also given that, a₁₀ = 21 ⟹ 21 = a + (10 – 1)d ⟹ 21 = a + 9d …. (2) Subtracting (2) from (1), we get 24 – 21 = 2a + 9d – a – 9d ⟹a = 3 Now, on putting a = 3 in equation (2), we get 3 + 9d = 21 9d = 18 d = 2 Thus, we have the first term(a) = 3 and the common difference(d) = 2 Therefore, the n^th term is given by a_n = a + (n – 1)d = 3 + (n – 1)2 = 3 + 2n -2 = 2n + 1 Hence, the n^th term of the A.P is (a_n) = 2n + 1.

If the 10th term of an A.P. is 21 and the sum of its first 10 terms is 120, find its nth term.

Answer»

Let’s consider a to be the first term and d be the common difference.

And we know that, sum of first n terms is:

S_n = \(\frac{n}{2}\)(2a + (n − 1)d) and n^th term is given by: a_n = a + (n – 1)d

Now, from the question we have

S₁₀ = 120

⟹ 120 = \(\frac{10}{2}\)(2a + (10 − 1)d)

⟹ 120 = 5(2a + 9d)

⟹ 24 = 2a + 9d …. (1)

Also given that, a₁₀ = 21

⟹ 21 = a + (10 – 1)d

⟹ 21 = a + 9d …. (2)

Subtracting (2) from (1), we get

24 – 21 = 2a + 9d – a – 9d

⟹a = 3

Now, on putting a = 3 in equation (2), we get

3 + 9d = 21

9d = 18

d = 2

Thus, we have the first term(a) = 3 and the common difference(d) = 2

Therefore, the n^th term is given by

a_n = a + (n – 1)d = 3 + (n – 1)2

= 3 + 2n -2

= 2n + 1

Hence, the n^th term of the A.P is (a_n) = 2n + 1.