Advertisements
Advertisements
प्रश्न
If the sum of n terms of an A.P. is Sn = 3n2 + 5n. Write its common difference.
Advertisements
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 given below:
`7 + 10 1/2 + 14 + ... + 84`
Find the sum of first 15 multiples of 8.
If the nth term of the A.P. 9, 7, 5, ... is same as the nth term of the A.P. 15, 12, 9, ... find n.
Find the 12th term from the end of the following arithmetic progressions:
3, 5, 7, 9, ... 201
Find the sum of n terms of an A.P. whose nth terms is given by an = 5 − 6n.
The first term of an A.P. is 5, the last term is 45 and the sum of its terms is 1000. Find the number of terms and the common difference of the A.P.
The fourth term of an A.P. is 11 and the eighth term exceeds twice the fourth term by 5. Find the A.P. and the sum of first 50 terms.
The 8th term of an AP is zero. Prove that its 38th term is triple its 18th term.
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).
Find four consecutive terms in an A.P. whose sum is 12 and sum of 3rd and 4th term is 14.
(Assume the four consecutive terms in A.P. are a – d, a, a + d, a +2d)
The sum of first seven terms of an A.P. is 182. If its 4th and the 17th terms are in the ratio 1 : 5, find the A.P.
If the sum of three consecutive terms of an increasing A.P. is 51 and the product of the first and third of these terms is 273, then the third term is
If Sn denote the sum of the first n terms of an A.P. If S2n = 3Sn, then S3n : Sn is equal to
If 18th and 11th term of an A.P. are in the ratio 3 : 2, then its 21st and 5th terms are in the ratio
Q.10
Q.15
Q.20
How many terms of the A.P. 24, 21, 18, … must be taken so that the sum is 78? Explain the double answer.
If Sn denotes the sum of first n terms of an AP, prove that S12 = 3(S8 – S4)
Find the sum of first 17 terms of an AP whose 4th and 9th terms are –15 and –30 respectively.
