Advertisements
Advertisements
प्रश्न
If the numbers (2n – 1), (3n+2) and (6n -1) are in AP, find the value of n and the numbers
उत्तर
It is given that the numbers (2n-1) , (3n +2) and (6n -1) are in AP.
∴ (3n + 2) - (2n-1) = (6n-1) - (3n+2)
⇒ 3n + 2-2n +1 = 6n-1-3n-2
⇒ n +3=3n-3
⇒ 2n = 6
⇒ n = 3
When , n = 3
2n - 1 = 2×3 -1=6-1=5
3n + 2 = 3×3+2=9+2=11
6n -1 = 6 × 3-1=18-1=17
Hence, the required value of n is 3 and the numbers are 5, 11 and 17.
APPEARS IN
संबंधित प्रश्न
The 4th term of an AP is 11. The sum of the 5th and 7th terms of this AP is 34. Find its common difference
If an denotes the nth term of the AP 2, 7, 12, 17, … find the value of (a30 - a20 ).
In an A.P. sum of three consecutive terms is 27 and their product is 504, find the terms.
(Assume that three consecutive terms in A.P. are a – d, a, a + d).
The sum of the first n terms of an A.P. is 3n2 + 6n. Find the nth term of this A.P.
The sum of first n terms of an A.P. is 3n2 + 4n. Find the 25th term of this A.P.
The first three terms of an A.P. respectively are 3y − 1, 3y + 5 and 5y + 1. Then, y equals
Q.2
The sum of the first three terms of an Arithmetic Progression (A.P.) is 42 and the product of the first and third term is 52. Find the first term and the common difference.
Find the sum of three-digit natural numbers, which are divisible by 4
An AP consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three is 429. Find the AP.
