Advertisements
Advertisements
Question
The sum of first n terms of an A.P. is 5n − n2. Find the nth term of this A.P.
Solution
Let a be the first term and d be the common difference.
We know that, sum of first n terms = Sn = \[\frac{n}{2}\][2a + (n − 1)d] It is given that sum of the first n terms of an A.P. is 5n − n2.
∴ First term = a = S1 = 5(1) − (1)2 = 4.
Sum of first two terms = S2 = 5(2) − (2)2 = 6.
∴ Second term = S2 − S1 = 6 − 4 = 2.
∴ Common difference = d = Second term − First term
= 2 − 4 = −2
Also, nth term = an = a + (n − 1)d
⇒ an = 4 + (n − 1)(−2)
⇒ an = 4 − 2n + 2
⇒ an = 6 − 2n
Thus, nth term of this A.P. is 6 − 2n.
APPEARS IN
RELATED QUESTIONS
If Sn1 denotes the sum of first n terms of an A.P., prove that S12 = 3(S8 − S4).
How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?
Find the sum of n terms of an A.P. whose nth terms is given by an = 5 − 6n.
The sum of three consecutive terms of an AP is 21 and the sum of the squares of these terms is 165. Find these terms
Find the first term and common difference for the A.P.
5, 1, –3, –7,...
If the sum of n terms of an A.P. is 2n2 + 5n, then its nth term is
In an A.P. (with usual notations) : given a = 8, an = 62, Sn = 210, find n and d
How many terms of the A.P. 25, 22, 19, … are needed to give the sum 116 ? Also find the last term.
If an = 3 – 4n, show that a1, a2, a3,... form an AP. Also find S20.
Shubhankar invested in a national savings certificate scheme. In the first year he invested ₹ 500, in the second year ₹ 700, in the third year ₹ 900 and so on. Find the total amount that he invested in 12 years