Advertisements
Advertisements
प्रश्न
If sum of first 6 terms of an AP is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.
Advertisements
उत्तर
Let a and d be the first term and common difference, respectively of an AP.
∵ Sum of n terms of an AP,
Sn = `n/2 [2a + (n - 1)d]` ...(i)
Now, S6 = 36 ...[Given]
⇒ `6/2[2a + (6 - 1)d]` = 36
⇒ 2a + 5d = 12 ...(ii)
And S16 = 256
⇒ `16/2[2a + (16 - 1)d]` = 256
⇒ 2a + 15d = 32 ...(iii)
On subtracting equation (ii) from equation (iii), we get
10d = 20
⇒ d = 2
From equation (ii),
2a + 5(2) = 12
⇒ 2a = 12 − 10 = 2
⇒ a = 1
∴ S10 = `10/2 [2a + (10 - 1)d]`
= 5[2(1) + 9(2)]
= 5(2 + 18)
= 5 × 20
= 100
Hence, the required sum of first 10 terms is 100.
APPEARS IN
संबंधित प्रश्न
How many terms of the A.P. 18, 16, 14, .... be taken so that their sum is zero?
The sum of n, 2n, 3n terms of an A.P. are S1 , S2 , S3 respectively. Prove that S3 = 3(S2 – S1 )
Find the sum of first 40 positive integers divisible by 6.
In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato and other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.
A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
[Hint: to pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 × (5 + 3)]
How many terms are there in the A.P. whose first and fifth terms are −14 and 2 respectively and the sum of the terms is 40?
Find the sum:
25 + 28 + 31 + ... + 100
Find the sum of the first 15 terms of the following sequences having nth term as yn = 9 – 5n.
Find the sum of the first 25 terms of an A.P. whose nth term is given by an = 2 − 3n.
Find the sum of two middle most terms of the AP `-4/3, -1 (-2)/3, ..., 4 1/3.`
The 9th term of an AP is –32 and the sum of its 11th and 13th terms is –94. Find the common difference of the AP.
Find the value of x for which the numbers (5x + 2), (4x – 1) and (x + 2) are in AP.
Find the sum of first n even natural numbers.
Write an A.P. whose first term is a and common difference is d in the following.
a = 6, d = –3
There are 25 rows of seats in an auditorium. The first row is of 20 seats, the second of 22 seats, the third of 24 seats, and so on. How many chairs are there in the 21st row ?
The 19th term of an A.P. is equal to three times its sixth term. If its 9th term is 19, find the A.P.
Find the sum: 1 + 3 + 5 + 7 + ... + 199 .
The sum of first 9 terms of an A.P. is 162. The ratio of its 6th term to its 13th term is 1 : 2. Find the first and 15th term of the A.P.
The sum of first n terms of an A.P. is 3n2 + 4n. Find the 25th term of this A.P.
Write 5th term from the end of the A.P. 3, 5, 7, 9, ..., 201.
If the sum of n terms of an A.P. be 3n2 + n and its common difference is 6, then its first term is ______.
If the sum of n terms of an A.P. is 2n2 + 5n, then its nth term is
Q.1
Q.2
Find the sum of all members from 50 to 250 which divisible by 6 and find t13.
Find the sum:
`4 - 1/n + 4 - 2/n + 4 - 3/n + ...` upto n terms
Yasmeen saves Rs 32 during the first month, Rs 36 in the second month and Rs 40 in the third month. If she continues to save in this manner, in how many months will she save Rs 2000?
Show that the sum of an AP whose first term is a, the second term b and the last term c, is equal to `((a + c)(b + c - 2a))/(2(b - a))`
The nth term of an Arithmetic Progression (A.P.) is given by the relation Tn = 6(7 – n)..
Find:
- its first term and common difference
- sum of its first 25 terms
Which expression gives the sum of the first n natural numbers \[1 + 2 + \cdots + n\]?
