How many terms of the series 1 + 2 + 3 +...... will sum to 5050?
Solution:
I Method:
The given series is an AP and sum of n terms of AP = Sn = n/2[2a + (n − 1) × d]
First term (a) =1
Difference (d) = 2-1=1
Given -
Sum of total terms Sn = 5050
Number of terms n=?
Sn = n/2[2a + (n − 1) × d]
5050 = n/2[2*1 + (n − 1) × 1]
5050 = n/2[2 + n − 1]
5050 = n/2[n+1]
10100 = n[n+1]
10100 = n2+n
n2+n -10100 = 0
n2 +(101-100)n -10100 = 0
n2 + 101n - 100n -10100 = 0
n (n+101) - 100(n+101) = 0
(n-100) (n+101) = 0
n = 100 / n = -101
The value of n cannot be negative, so n = 100
So, the sum of 100 terms of the series 1 + 2 + 3 +..... will be 5050.
II Method:
Sum of n natural numbers = n(n+1)/2
According to the question -
5050 = n(n+1)/2
10100 = n2+n
n2+n = 10100
n2 +(101-100)n -10100 = 0
n2 + 101n - 100n -10100 = 0
n (n+101) - 100(n+101) = 0
(n-100) (n+101) = 0
n = 100 / n = -101
The value of n cannot be negative, so n = 100
So, the sum of 100 terms of the series 1 + 2 + 3 +..... will be 5050.
Hence the correct answer is option C.
हल:
I Method:
दी गयी श्रेणी एक सामानांतर श्रेणी है और सामानांतर श्रेणी के n पदों का योग = Sn= n/2[2a + (n − 1) × d]
प्रथम पद a =1
सर्वान्तर d = 2-1=1
कुल पदों का योग Sn = 5050
पदों की संख्या n=?
Sn = n/2[2a + (n − 1) × d]
5050 = n/2[2*1 + (n − 1) × 1]
5050 = n/2[2 + n − 1]
5050 = n/2[n+1]
10100 = n[n+1]
10100 = n2+n
n2 +n -10100 = 0
n2 +(101-100)n -10100 = 0
n2 + 101n - 100n -10100 = 0
n (n+101) - 100(n+101) = 0
(n-100) (n+101) = 0
n = 100 / n = -101
n का मान ऋणात्मक नहीं हो सकता है अतः n = 100
अतः श्रेणी 1 + 2 + 3 +...... के 100 पदों का योग 5050 होगा l
II Method:
n प्राकृतिक संख्याओं का योग = n(n+1)/2
प्रशानानुसार -
5050 = n(n+1)/2
10100 = n2+n
n2+n = 10100
n2 +(101-100)n -10100 = 0
n2 + 101n - 100n -10100 = 0
n (n+101) - 100(n+101) = 0
(n-100) (n+101) = 0
n = 100 / n = -101
n का मान ऋणात्मक नहीं हो सकता है अतः n = 100
अतः श्रेणी 1 + 2 + 3 +...... के 100 पदों का योग 5050 होगा l
अतः सही उत्तर विकल्प C है l
