Advertisements
Advertisements
Question
The sum of first n terms of an AP is 3n2 + 4n. Find the 25th term of this AP ?
Solution
The sum of first n terms (Sn) is given as `S_n=3n^2+4n`.
So, first term (a1) =`S_n=3(1)^2+4(1)=7`
`S_2=a_1+a_2=3(2)^2+4(1)=7`
a2 = 20 − a1 = 20 − 7 = 13
So, common difference (d) = a2− a1 = 13 − 7 = 6
∴nth term is given by, an = a + (n − 1) d
Thus, 25th term = a25 = 7 + (25 − 1) × 6 = 7 + 24 × 6 = 7 + 144 = 151
APPEARS IN
RELATED QUESTIONS
In the following APs, find the missing terms in the boxes:
`square, 38, square, square, square, -22`
The first term of an A.P. is 5 and its 100th term is -292. Find the 50th term of this A.P.
The third term of an A.P. is 7 and the seventh term exceeds three times the third term by 2. Find the first term, the common difference and the sum of first 20 terms.
Show that each of the progressions given below is an AP. Find the first term, common difference and next term of each
(ii) 11, 6, 1, – 4,……..
Find:
(ii) the 35th term of AP 20,17,14,11,..........
Find:
the 18th term of the AP `sqrt(2), sqrt (18), sqrt(50), sqrt(98)`,...........
The 6th term from the end of the AP: 5, 2, -1, -4, …., -31, is ______.
The number of multiples lie between n and n2 which are divisible by n is ______.
How many three-digit numbers are divisible by 7?
Fill in the blank in the following table, given that a is the first term, d the common difference, and an nth term of the AP:
| a | d | n | an |
| -18 | ______ | 10 | 0 |
