Question

The sum of the n consecutive natural odd numbers starting from 3 is 35. find the value of n.

Answer 1

the n consecutive natural odd numbers starting from 3 are, 3 , 5 , 7 , n-terms is 35. these are in A.P.

where, a = 3 , d = 5 - 3 = 2 , Sn = 35.

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

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

35 * 2 = n[6 + 2n - 2]

70 = n(4 + 2n)

70 = 4n + 2n²

n² + 2n - 35 = 0

n² + 7n - 5n - 35 = 0

n(n + 7) - 5(n + 7) = 0

(n - 5)(n + 7) = 0

n = 5 , n = -7[NEGLECT]

so, n = 5

Answer 2

Given,
the sum of n consecutive natural odd numbers starting from 3 is 35.
Basically, series is
3,5,7,9........and so on upto nth terms.
Now the following series is in AP
Then,
First term(a) = 3
Common difference (d)= 5-3= 2
number of terms= n
Sum of nth terms (Sn)= 35
We know that sum of nth terms of AP is given as:-
Sn= n/2[2a + (n-1)d]
35= n/2 [2×3 + (n-1)*2]
35= n/2[ 6+ 2n-2]
35= n/2[4+2n]
35= n[2+n]
35= 2n + n²
2n+n²-35= 0
n²+2n-35= 0
n² 7n-5n-35=0
n(n+7) -5(n+7)= 0
(n+7)(n-5)= 0
n≠-7 ( as numbers of terms can't be negative)
So,
n= 5.


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