Question

The sum of the n consecutive natural numbers starting from 3 is 42. Find the value of n.

Answer 1

Hi Friend !!!

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According to the question,
a = 3.
d = 1


Sn = n/2[2a+(n-1)d]

42 = n/2[2(3)+(n-1)1]

42×2 = n(6+n-1)

84 = n(n+5)

84 = n²+5n

n²+5n-84 = 0

n²+12n-7n-84 = 0

n(n+12)-7(n+12) = 0

(n-7)(n+12) = 0

n= -12 (not possible) or n= 7

So n = 7


here is ur answer

Answer 2

Sol : Since, the consecutive natural number form an A.P.

Given,

First term ( a ) = 3

Common difference ( d ) = 1

Sum of ' n ' terms  ( s ) = 42.

Number of terms ( n ) = ?

Using formula for sum of an A.P.,

 s = ( n / 2 ) { 2 a + ( n - 1 ) d }

By substituting the values of 's' , 'a' and 'd',

42 = ( n / 2 ) { 2 x 3 + ( n - 1 ) 1 }

42 = ( n / 2 ) { 6 + n - 1 }

42 = ( n / 2 ) ( 5 + n )

42 x 2 = n ( 5 + n )

84 = 5 n + n²

n² + 5 n - 84 = 0

n² + 12 n - 7 n - 84 = 0

n ( n + 12 ) - 7 ( n + 12 ) = 0

( n + 12 ) ( n - 7 ) = 0

( n + 12 ) = 0   or ( n - 7 ) = 0

 n = - 12          or n = 7.

Since number of terms of any A.P. can't be zero.So , number of terms = 7.

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