How many Handshakes!!!

as we know the sum of n consecutive nos= n*(n+1)/2

putting equals to 66

=> n=11

now 1 man can handshake 11 people

so 11+1=12 people

$\text{number of handshakes}=\dfrac{n}{2}(n-1)$

$66=\dfrac{n}{2}(n-1)$

$132=n^2-n$

$n^2-n-132=0$

$(n-12)(n+11)=0$

$n=12$ or $n=-11$

Reject the negative value. Therefore, there were 12 people at the party.

there are n people then the first person shook hands with (n-1) people then the second person (n-2) people to shake hands with coz he already shook hands with the first person. similarly third person can shake hands with (n-3) people. this continues till there is one handshake left . so the sum of all these handshakes is equal to 66. => (n-1) +(n-2)+(n-3)+.......+2+1=66 => (n-1)/2 (n-1+1)=66 => n (n-1)=66 2=2 2 3 11=12*11 comparing both sides we get, n=12