the sum of first n even natural numbers is 325 find n
Answers
Let S be the required sum.
Therefore, S = 1 + 2 + 3 + 4 + 5 + .................... + 25
Clearly, it is an Arithmetic Progression whose first term = 1, last term = 25 and number of terms = 25.
Therefore, S = 252(25 + 1), [Using the formula S = n2(a + l)]
= 252(26)
= 25 × 13
= 325
Therefore, the sum of first 25 natural numbers is 325.
Sum of 25 natural numbers is 325
Given : sum of n natural numbers is 325.
To Find : Value of n
Solution :
natural numbers are counting numbers 1 ,2 , 3
Arithmetic sequence
Sequence of terms in which difference between one term and the next is a constant.
This is also called Arithmetic Progression AP
Arithmetic sequence can be represented in the form :
a, a + d , a + 2d , …………………………, a + (n-1)d
a = First term
d = common difference = aₙ-aₙ₋₁
nth term = aₙ = a + (n-1)d
Sₙ = (n/2)(2a + (n - 1)d)
Sum of n natural numbers
(n/2)(2 * 1 + (n - 1) * 1)
= (n/2)(2 + n - 1)
= (n )(n + 1)/2
n(n + 1)/2 = 325
=> n² + n - 650 = 0
=> n² + 26n - 25n - 650 =0
=> n(n + 26) - 25(n + 26) = 0
=> (n - 25)(n + 26) = 0
=> n = 25 , n = - 26
n can not be negative
Hence Value of n is 25
Sum of 25 natural numbers is 325
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