Advertisements
Advertisements
प्रश्न
In an AP, if Sn = n(4n + 1), find the AP.
उत्तर
We know that, the nth term of an AP is
an = Sn – Sn – 1
an = n(4n + 1) – (n – 1){4(n – 1) + 1} ...[∵ Sn = n(4n + 1)]
⇒ an = 4n2 + n – (n – 1)(4n – 3)
= 4n2 + n – 4n2 + 3n + 4n – 3
= 8n – 3
Put n = 1,
a1 = 8(1) – 3
= 5
Put n = 2,
a2 = 8(2) – 3
= 16 – 3
= 13
Put n = 3,
a3 = 8(3) – 3
= 24 – 3
= 21
Hence, the required AP is 5, 13, 21,...
APPEARS IN
संबंधित प्रश्न
In an AP given a = 8, an = 62, Sn = 210, find n and d.
The sum of the first n terms of an AP is (3n2+6n) . Find the nth term and the 15th term of this AP.
The first and the last terms of an A.P. are 8 and 350 respectively. If its common difference is 9, how many terms are there and what is their sum?
The first and last terms of an A.P. are 1 and 11. If the sum of its terms is 36, then the number of terms will be
Let Sn denote the sum of n terms of an A.P. whose first term is a. If the common difference d is given by d = Sn − kSn−1 + Sn−2, then k =
The sum of n terms of two A.P.'s are in the ratio 5n + 9 : 9n + 6. Then, the ratio of their 18th term is
If \[\frac{5 + 9 + 13 + . . . \text{ to n terms} }{7 + 9 + 11 + . . . \text{ to (n + 1) terms}} = \frac{17}{16},\] then n =
Obtain the sum of the first 56 terms of an A.P. whose 18th and 39th terms are 52 and 148 respectively.
Find the common difference of an A.P. whose first term is 5 and the sum of first four terms is half the sum of next four terms.
Find t21, if S41 = 4510 in an A.P.