Advertisements
Advertisements
प्रश्न
Find the sum: 1 + 3 + 5 + 7 + ... + 199 .
Advertisements
उत्तर
1 + 3 + 5 + 7 + ... + 199 .
Common difference of the A.P. (d) = `a_2 - a
_1`
= 3-1
= 2
So here,
First term (a) = 1
Last term (l) = 199
Common difference (d) = 2
So, here the first step is to find the total number of terms. Let us take the number of terms as n.
Now, as we know,
`a_n = a + ( n - 1)d`
So, for the last term,
199 = 1 + (n-1)2
199 = 1 + 2n - 2
199+1 = 2n
n = `200/2`
n = 100
Now, using the formula for the sum of n terms, we get
`S_n = 100/2 [2(1) + (100 - 1) 2 ]`
=50 [ 2 + (99) 2]
= 50 (2 + 198)
On further simplification, we get,
`S_n = 50(200)`
= 10000
Therefore, the sum of the A.P is `S_n = 10000 `
APPEARS IN
संबंधित प्रश्न
Find the sum of the following arithmetic progressions:
41, 36, 31, ... to 12 terms
Find the sum of the first 25 terms of an A.P. whose nth term is given by an = 2 − 3n.
If the pth term of an AP is q and its qth term is p then show that its (p + q)th term is zero
How many two-digit number are divisible by 6?
The sum of three numbers in AP is 3 and their product is -35. Find the numbers.
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 sum of the first n terms of an AP in `((5n^2)/2 + (3n)/2)`.Find its nth term and the 20th term of this AP.
If the ratio of sum of the first m and n terms of an AP is m2 : n2, show that the ratio of its mth and nth terms is (2m − 1) : (2n − 1) ?
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)
Fill up the boxes and find out the number of terms in the A.P.
1,3,5,....,149 .
Here a = 1 , d =b`[ ], t_n = 149`
tn = a + (n-1) d
∴ 149 =`[ ] ∴149 = 2n - [ ]`
∴ n =`[ ]`
Let there be an A.P. with first term 'a', common difference 'd'. If an denotes in nth term and Sn the sum of first n terms, find.
Which term of the sequence 114, 109, 104, ... is the first negative term?
Find the sum of first 20 terms of an A.P. whose first term is 3 and the last term is 57.
If the third term of an A.P. is 1 and 6th term is – 11, find the sum of its first 32 terms.
If an = 3 – 4n, show that a1, a2, a3,... form an AP. Also find S20.
Find the sum of numbers between 1 to 140, divisible by 4.
The sum of first 16 terms of the AP: 10, 6, 2,... is ______.
The sum of first five multiples of 3 is ______.
Solve the equation
– 4 + (–1) + 2 + ... + x = 437
Find the sum of first 25 terms of the A.P. whose nth term is given by an = 5 + 6n. Also, find the ratio of 20th term to 45th term.
