Question

श्रेणी 1+2+3.... के कितने पदों का योगफल 5050 होगा ?​

Answer 1

sum of natural numbers= n×(n+1)/2=5050

n^2+n-10100=0

n^2+101n-100n=0

n=100 or -101

n can't be -101 because ek shrebi mein rinatmak pad NAHI HO SKTE....

so n=100

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Answer 2

The required number of terms is 100.

Step-by-step explanation:

We need to find number of terms in the series 1+2+3+.., whose sum is 5050.

First term = 1

Common difference = 2-1 =1

Sum n terms of an AP is

$S_n=\dfrac{n}{2}[2a+(n-1)d]$

where, a is first term and d is common difference.

Substitute a=1, d=1 and $S_n=5050$

$5050=\dfrac{n}{2}[2(1)+(n-1)(1)]$

$10100=n[2+n-1]$

$10100=n(1+n)$

$10100=n^2+n$

$n^2+n-10100=0$

Splitting the middle term we get

$n^2+101n-100n-10100=0$

$n(n+101)-100(n+101)=0$

$(n+101)(n-100)=0$

$n=-101,100$

Number of terms can not be negative. So, n=100.

#Learn more

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