Advertisements
Advertisements
प्रश्न
Find the sum of all even numbers from 1 to 250.
Advertisements
उत्तर
The even numbers from 1 to 250 are 2, 4, 6, 8, ............, 250
And the last even number is = 250
∴ an = 250, a = 2 and d = 2
∴ an = a + (n – 1)d
∴ 250 = 2 + (n – 1)2
∴ 250 = 2 + 2n – 2
∴ 2n = 250
∴ n = `250/2` = 125
From 1 to 250, there are 125 even numbers.
Since, we know `S_n = n/2 [a + a_n]`
∴ The sum of 125 even numbers
`S_125 = 125/2 [2 + 250]`
= `125/2 xx 252`
= 125 × 126
= 15,750
As a result, the total of all even numbers from 1 to 250 is 15,750.
APPEARS IN
संबंधित प्रश्न
If the sum of the first n terms of an A.P. is `1/2`(3n2 +7n), then find its nth term. Hence write its 20th term.
Find the sum of first 20 terms of the following A.P. : 1, 4, 7, 10, ........
The ratio of the sum use of n terms of two A.P.’s is (7n + 1) : (4n + 27). Find the ratio of their mth terms
Find the sum given below:
–5 + (–8) + (–11) + ... + (–230)
Find the sum of the first 15 terms of each of the following sequences having the nth term as
`a_n = 3 + 4n`
Find the 6th term form the end of the AP 17, 14, 11, ……, (-40).
The first and last terms of an AP are a and l respectively. Show that the sum of the nth term from the beginning and the nth term form the end is ( a + l ).
Find the value of x for which the numbers (5x + 2), (4x - 1) and (x + 2) are in AP.
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 =`[ ]`
The sum of the first n terms of an A.P. is 3n2 + 6n. Find the nth term of this A.P.
If 18, a, b, −3 are in A.P., the a + b =
The common difference of the A.P. \[\frac{1}{2b}, \frac{1 - 6b}{2b}, \frac{1 - 12b}{2b}, . . .\] is
The term A.P is 8, 10, 12, 14,...., 126 . find A.P.
A manufacturer of TV sets produces 600 units in the third year and 700 units in the 7th year. Assuming that the production increases uniformly by a fixed number every year, find:
- the production in the first year.
- the production in the 10th year.
- the total production in 7 years.
Show that a1, a2, a3, … form an A.P. where an is defined as an = 3 + 4n. Also find the sum of first 15 terms.
Find the sum of first 1000 positive integers.
Activity :- Let 1 + 2 + 3 + ........ + 1000
Using formula for the sum of first n terms of an A.P.,
Sn = `square`
S1000 = `square/2 (1 + 1000)`
= 500 × 1001
= `square`
Therefore, Sum of the first 1000 positive integer is `square`
Find the sum of natural numbers between 1 to 140, which are divisible by 4.
Activity: Natural numbers between 1 to 140 divisible by 4 are, 4, 8, 12, 16,......, 136
Here d = 4, therefore this sequence is an A.P.
a = 4, d = 4, tn = 136, Sn = ?
tn = a + (n – 1)d
`square` = 4 + (n – 1) × 4
`square` = (n – 1) × 4
n = `square`
Now,
Sn = `"n"/2["a" + "t"_"n"]`
Sn = 17 × `square`
Sn = `square`
Therefore, the sum of natural numbers between 1 to 140, which are divisible by 4 is `square`.
In a ‘Mahila Bachat Gat’, Sharvari invested ₹ 2 on first day, ₹ 4 on second day and ₹ 6 on third day. If she saves like this, then what would be her total savings in the month of February 2010?
Find the sum of odd natural numbers from 1 to 101
Find the sum:
`(a - b)/(a + b) + (3a - 2b)/(a + b) + (5a - 3b)/(a + b) +` ... to 11 terms
