Advertisements
Advertisements
प्रश्न
If the sum of n terms of an A.P. is Sn = 3n2 + 5n. Write its common difference.
उत्तर
Here, we are given,
`S_n = 3n^2 + 5n`
Let us take the first term as a and the common difference as d.
Now, as we know,
`a_n = S_n - S_(n-1)`
So, we get,
`a_n = (3n^^^^2 + 5n) - [3(n-1)^2 + 5 (n-1)]`
`=3n^2 + 5n - [3(n^2 + 1 - 2n) + 5n - 5] [\text{ Using} (a - b)^2= a^2 - ab]`
`=3n^2 + 5n - (3n^2 + 3 - 6n + 5n - 5)`
`=3n^2 + 5n - 3n^2 - 3 + 6n - 5n + 5`
= 6n + 2 ..................(1)
Also,
`a_n = a + (n-1)d`
= a + nd - d
= nd + ( a- d) ...............(2)
On comparing the terms containing n in (1) and (2), we get,
dn = 6n
d = 6
Therefore, the common difference is d = 6 .
APPEARS IN
संबंधित प्रश्न
Find the sum of first 40 positive integers divisible by 6.
Which term of AP 72,68,64,60,… is 0?
Find the 8th term from the end of the AP 7, 10, 13, ……, 184.
How many two-digit number are divisible by 6?
If the sum of first n terms is (3n2 + 5n), find its common difference.
Q.16
Q.20
Find the sum of all members from 50 to 250 which divisible by 6 and find t13.
Find the sum of first 1000 positive integers.
Activity :- Let 1 + 2 + 3 + ........ + 1000
Using formula for the sum of first n terms of an A.P.,
Sn = `square`
S1000 = `square/2 (1 + 1000)`
= 500 × 1001
= `square`
Therefore, Sum of the first 1000 positive integer is `square`
Measures of angles of a triangle are in A.P. The measure of smallest angle is five times of common difference. Find the measures of all angles of a triangle. (Assume the measures of angles as a, a + d, a + 2d)