Question

# If ( S_{n}=n P+frac{n}{2}(n-1) Q, ) where ( S_{n} ) denotes the sum of the first ( n ) terms of an A.P., then the common difference is

Solution

The formula for ( S_{n}=frac{n}{2}[2 a+(n-1) d]-0 )

Given eq, Sn = nP + (frac{n}{2}(n-1) Q)

= (frac{2np}{2}) + (frac{n}{2})(n-1 Q)

= (frac{n}{2}[2P + (n - 1) Q])

Company with 1, P = a, d = Q

( therefore ) common diffence ( Q )