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abby_g
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1. The sum of the even numbers between 1 and n is 79*80, where n is an odd number, then n=?
Sol: First term a=2, common difference d=2 since even number
therefore sum to first n numbers of Arithmetic progression would be
n/2(2a+(n-1)d)
= n/2(2*2+(n-1)*2)=n(n+1) and this is equal to 79*80
therefore n=79 which is odd...
I think the answer is 159.
we have the series between 1 to n as 2,4,6,8,...,n-1
so take 2 common and we have
2[1,2,3,...,(n-1)/2]
:arrow: the sum = 2[ (n-1)/2 * {(n-1)/2 + 1)]/2
:arrow: = (n-1)(n+1)/4
:arrow: equating on bothsides (n-1)(n+1)/4 = 79 *80
:arrow: (n-1)(n+1) = 158 * 160
:arrow: Hence n=159
The original method mistooks n(number of terms) for n (of 1 to n ) where n is number and not number of terms. Can anybody correct me please.
Sol: First term a=2, common difference d=2 since even number
therefore sum to first n numbers of Arithmetic progression would be
n/2(2a+(n-1)d)
= n/2(2*2+(n-1)*2)=n(n+1) and this is equal to 79*80
therefore n=79 which is odd...
I think the answer is 159.
we have the series between 1 to n as 2,4,6,8,...,n-1
so take 2 common and we have
2[1,2,3,...,(n-1)/2]
:arrow: the sum = 2[ (n-1)/2 * {(n-1)/2 + 1)]/2
:arrow: = (n-1)(n+1)/4
:arrow: equating on bothsides (n-1)(n+1)/4 = 79 *80
:arrow: (n-1)(n+1) = 158 * 160
:arrow: Hence n=159
The original method mistooks n(number of terms) for n (of 1 to n ) where n is number and not number of terms. Can anybody correct me please.












