plz. help friends to solve this

2) the mean and s.d of a b and 2 are 3 and 1 respectively.then what is the value of ab

given, mean = 3

implies that

(a+b+2)/3=3

so, a+b+2=3*3

a+b=7  --------------------  eqn (1)

given, s.d =1, implies variance = 1*1

i.e, variance, (a-m)²+(b-m)²+(2-m)² = 1 ; here m, mean =3

{(a-3) ²+(b-3) ²+(2-3) ² }/3= 1

a² -6a+9+ b² -6b+9+1=3

a² + b² -6(a+b)=3-19

a² + b² -6*7=3-19              from eqn (1)

a² + b² = 26 ……. (42-16)

(a+b)² = a² + b²+2ab

therefore; 2ab = 7 ² -26

ab = 23/2

1.if the s.d. of the first n natural no. is 2, then what is the value of n?

n=7

[{1 – (n+1)/2} ² +{2 – (n+1)/2} ² +{3 – (n+1)/2} ² +…….+{n – (n+1)/2} ²  ]/n = 4

[1²  +2²  +3²  +….+ n² ] + 2(n+1)/2{1+2+3+…+n}+ n*(n+1)²/2²              =4n

n(n+1)(2n+1)/6     –    {2(n+1)/2}*n(n+1)/2   +    n(n+1)²/4         =4n

n(n+1)/2  { (2n+1)/3- (n+1)/2 + (n+1)/2 }                       =4n

(n+1) {4n+2-3n-3}  /   6                   =4*2

(n+1) (n-1)                     =8*6

n²  -   1          =48

therefore, n=7

1. f the s.d. of the first n natural no. is 2, then what is the value of n?