Advertisements
Advertisements
Question
The sum of the first n terms of an A.P. is 3n2 + 6n. Find the nth term of this A.P.
Advertisements
Solution
Given: Sn = 3n2 + 6n
We know that the nth term of an A.
∴ an = Sn − Sn−1
Sn−1 = 3(n − 1)2 + 6(n − 1)
= 3(n2 − 2n + 1) + 6n − 6
= 3n2 − 6n + 3 + 6n − 6
= 3n2 − 3
∴ an = Sn − Sn−1
= (3n2 + 6n) − (3n2 − 3)
= 3n2 + 6n − 3n2 + 3
= 6n + 3
Thus, nth term of this A.P. is 6n + 3.
APPEARS IN
RELATED QUESTIONS
The ratio of the sums of m and n terms of an A.P. is m2 : n2. Show that the ratio of the mth and nth terms is (2m – 1) : (2n – 1)
How many multiples of 4 lie between 10 and 250?
In an AP given d = 5, S9 = 75, find a and a9.
How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?
Show that a1, a2,..., an... form an AP where an is defined as below:
an = 9 − 5n
Also, find the sum of the first 15 terms.
The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceed the second term by 6, find three terms.
Find the 6th term form the end of the AP 17, 14, 11, ……, (-40).
Find the three numbers in AP whose sum is 15 and product is 80.
Find the sum of first n even natural numbers.
Find the sum of first n terms of an AP whose nth term is (5 - 6n). Hence, find the sum of its first 20 terms.
If the sum of first p terms of an A.P. is equal to the sum of first q terms then show that the sum of its first (p + q) terms is zero. (p ≠ q)
The sum of the first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28th term of this A.P.
Mark the correct alternative in each of the following:
If 7th and 13th terms of an A.P. be 34 and 64 respectively, then its 18th term is
If \[\frac{1}{x + 2}, \frac{1}{x + 3}, \frac{1}{x + 5}\] are in A.P. Then, x =
If k, 2k − 1 and 2k + 1 are three consecutive terms of an A.P., the value of k is
The sum of first 15 terms of an A.P. is 750 and its first term is 15. Find its 20th term.
How many terms of the A.P. 25, 22, 19, … are needed to give the sum 116 ? Also find the last term.
Find second and third terms of an A.P. whose first term is – 2 and the common difference is – 2.
Find the sum:
`4 - 1/n + 4 - 2/n + 4 - 3/n + ...` upto n terms
In an A.P., if Sn = 3n2 + 5n and ak = 164, find the value of k.
