Advertisements
Advertisements
प्रश्न
If the sum of the first four terms of an AP is 40 and that of the first 14 terms is 280. Find the sum of its first n terms.
उत्तर
Given that S4 = 40 and S14 = 280'
`"S"_"n" = "n"/2[2"a" + ("n-1)d"]`
`"S"_4 = 4/2[2"a" + (4-1)"d"] = 40`
`=> 2"a" + 3"d" = 20` .......(i)
`"S"_14 = 14/2 [2"a" + (14 -1)"d"] = 280`
`=> 2"a" + 13"d" = 40` ...(ii)
(ii) - (i)
10d = 20 ⇒ d = 2
Sunstituting the value of d in (i) we get
2a + 6 = 20 ⇒ a = 7
Sum of first n terms,
`"S"_"n" = "n"/2[2"a" + ("n-1)d"]`
= `"n"/2 [14 + ("n"-1) 2]`
= n ( 7 + n - 1)
= n (n + 6)
= n2 + 6n
Therefore, Sn = n2 + 6n
संबंधित प्रश्न
Show that a1, a2,..., an... form an AP where an is defined as below:
an = 3 + 4n
Also, find the sum of the first 15 terms.
200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed, and how many logs are in the top row?
Find the sum of the following arithmetic progressions:
`(x - y)/(x + y),(3x - 2y)/(x + y), (5x - 3y)/(x + y)`, .....to n terms
Determine the A.P. Whose 3rd term is 16 and the 7th term exceeds the 5th term by 12.
Find out the sum of all natural numbers between 1 and 145 which are divisible by 4.
In an A.P., the sum of first ten terms is −150 and the sum of its next ten terms is −550. Find the A.P.
A man is employed to count Rs 10710. He counts at the rate of Rs 180 per minute for half an hour. After this he counts at the rate of Rs 3 less every minute than the preceding minute. Find the time taken by him to count the entire amount.
The common difference of the A.P. \[\frac{1}{2b}, \frac{1 - 6b}{2b}, \frac{1 - 12b}{2b}, . . .\] is
Q.4
If an = 3 – 4n, show that a1, a2, a3,... form an AP. Also find S20.