Question: If S<sub>n</sub> = np + $\frac { 1 } { 2 }$ n(n – 1)Q where S<sub>n</sub> = sum of first n terms ...

Question

If S_n = np + $\frac { 1 } { 2 }$ n(n – 1)Q where S_n = sum of first n terms of an A.P. then common difference =

Answer 1

Solution

S_n = $\mathrm { n } ^ { 2 } \cdot \left( \frac { \mathrm { a } } { 2 } \right) + \mathrm { n } \left( \mathrm { p } - \frac { \mathrm { Q } } { 2 } \right)$ ̃ C.d = $2 \times \left( \frac { \mathrm { Q } } { 2 } \right)$ = Q

Question: If S<sub>n</sub> = np + \(\frac { 1 } { 2 }\) n(n – 1)Q where S<sub>n</sub> = sum of first n terms ...

Solution