Question:
In an AP it is given that $S_{n}=q n^{2}$ and $S_{m}=q m^{2}$. Prove that $S_{q}=q^{3}$.
Solution:
Given: $S_{n}=q n^{2}, S_{m}=q m^{2}$
To prove: $\mathrm{S}_{\mathrm{q}}=\mathrm{q}^{3}$
Put $n=1$ we get
$a=q \ldots \ldots$ equation 1
Put n = 2
2a + d = 4q ……equation 2
Using equation 1 and 2 we get
d = 2q
So $\mathrm{S}_{\mathrm{q}}=\frac{\mathrm{q}}{2}(2 \mathrm{q}+(\mathrm{q}-1) \times 2 \mathrm{q})$
$\mathrm{S}_{\mathrm{q}}=\mathrm{q}^{3}$
Hence proved.