Advertisements
Advertisements
प्रश्न
In an A.P., the sum of first n terms is `n/2 (3n + 5)`. Find the 25th term of the A.P.
Advertisements
उत्तर
Given, Sn = `n/2 (3n + 5)`
∴ `S_(n - 1) = (n - 1)/2 [3(n - 1) + 5]`
or `S_(n - 1) = (n - 1)/2 (3n + 2)`
Since, an = Sn – Sn–1
= `n/2(3n + 5) - (n - 1)/2(3n + 2)`
= `(3n^2)/2 + (5n)/2 - (3n(n - 1))/2 - (2(n - 1))/2`
= `(3n^2)/2 + (5n)/2 - (3n^2)/2 + (3n)/2 - n + 1`
= `(8n)/2 - n + 1`
= 4n – n + 1
= 3n + 1
Now, a25 = 3(25) + 1
or, a25 = 75 + 1 = 76
Thus, 25th term of AP. is 76.
APPEARS IN
संबंधित प्रश्न
Find the sum of all natural numbers between 250 and 1000 which are exactly divisible by 3
In an AP given d = 5, S9 = 75, find a and a9.
Three numbers are in A.P. If the sum of these numbers is 27 and the product 648, find the numbers.
Find the A.P. whose fourth term is 9 and the sum of its sixth term and thirteenth term is 40.
Find out the sum of all natural numbers between 1 and 145 which are divisible by 4.
The 19th term of an A.P. is equal to three times its sixth term. If its 9th term is 19, find the A.P.
Q.1
Q.11
The sum of the first 2n terms of the AP: 2, 5, 8, …. is equal to sum of the first n terms of the AP: 57, 59, 61, … then n is equal to ______.
The sum of A.P. 4, 7, 10, 13, ........ upto 20 terms is ______.
