Advertisements
Advertisements
प्रश्न
The sum of n terms of an A.P. is 3n2. The second term of this A.P. is ______.
पर्याय
8
3
9
12
Advertisements
उत्तर
The sum of n terms of an A.P. is 3n2. The second term of this A.P. is 9.
Explanation:
Given Sn = 3n2;
Sn – 1 = 3(n – 1)2
∴ Tn = Sn – Sn–1
= 3n2 – 3(n – 1)2
= 3n2 – 3[n2 – 2n + 1] ...[∵ (a – b)2 = a2 – 2ab + b2]
= 3n2 – 3n2 + 6n – 3
= 6n – 3
∴ T2 = 6 × 2 – 3
= 12 – 3
= 9
APPEARS IN
संबंधित प्रश्न
Find the sum:
25 + 28 + 31 + ... + 100
Find the sum of first 12 natural numbers each of which is a multiple of 7.
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 7th term of an AP is –4 and its 13th term is –16. Find the AP.
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.
Write the next term of the AP `sqrt(8), sqrt(18), sqrt(32),`....
The nth term of an A.P. is given by (–4n + 15). Find the sum of first 20 terms of this A.P.
If Sn denotes the sum of first n terms of an A.P., prove that S12 = 3(S8 – S4).
The sum of first 14 terms of an A.P. is 1050 and its 14th term is 140. Find the 20th term.
How many terms of the A.P. 27, 24, 21, ..., should be taken so that their sum is zero?
