Asked by Olaitan
                How many terms of the sequence 1+2+3+4.... are needed to make a sum of 820
            
            
        Answers
                    Answered by
            Olaitan segun
            
    Sn=n/2 (2a + n-1)d
820=n/2(2(1)+(n-1)1)
820×2= n(2+(n-1)1)
1640= n(2+n-1)
1640=n(2-1+n)
1640= n(1+n)
1640= n+ n²
1640-n-n²=0
n²+n-1640=0
n²+41n-40n-1640 =0
n(n+41)-40(n+41)=0
n-40 =0 or n+41=0
n=40 or n= -41
That is, 14 terms are needed to make sum of 820
    
820=n/2(2(1)+(n-1)1)
820×2= n(2+(n-1)1)
1640= n(2+n-1)
1640=n(2-1+n)
1640= n(1+n)
1640= n+ n²
1640-n-n²=0
n²+n-1640=0
n²+41n-40n-1640 =0
n(n+41)-40(n+41)=0
n-40 =0 or n+41=0
n=40 or n= -41
That is, 14 terms are needed to make sum of 820
                    Answered by
            oobleck
            
    all that work, and you type the wrong answer at the end!
    
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