Question

If 10 times the 10th term of an A.P. is equal to 15 times the 15th term, show that 25th term of the A.P. is zero.

Answer 1

Given:
$$\begin{gathered} 10a_{10}=15a_{15} \\ \Rightarrow 10\left[a+\left(10-1\right)d\right]=15\left[a+\left(15-1\right)d\right] \\ \Rightarrow 10(a+9d)=15(a+14d) \\ \Rightarrow 10a+90d=15a+210d \\ \Rightarrow 0=5a+120d \\ \Rightarrow 0=a+24d \\ \Rightarrow a=-24d...\left(\mathrm{i}\right) \end{gathered}$$

To show:
$$\begin{gathered} a_{25}=0 \\ \Rightarrow \mathrm{LHS}:a_{25}=a+\left(25-1\right)d \\ =a+24d \\ =-24d+24d\left(\mathrm{From}\left(\mathrm{i}\right)\right) \\ =0=\mathrm{RHS} \\ \mathrm{Hence},\mathrm{proved}. \end{gathered}$$