Question

The fourth term of an ap is 0. Prove that the 25th term is 3 times the 11th term

Answer 1

$\bf\huge\textbf{\underline{\underline{According\:to\:the\:Question}}}$  

a4 = 0

a + 3d = 0

⇒ a = - 3d [Equation (1)]

an = a + (n - 1)d

a11 = a + 10d

Substitute the value of a from Equation (1)

= - 3d + 10d = 7d  

⇒ a25 = a + 24d

⇒ - 3d + 24d = 21d  (Equation 1)

= 3 × 7d

Therefore

a25 = 3 × a11

Answer 2

$\Large \text{Let} \\ \\ \textit{1st term = a \\ \\ 4th term = a_4 \\ \\ \textit{11th term}} = a_{11} \\ \\ \textit{25th term = }a_{25} \\ \\ \textit{common difference = d}$

$a_4=0 \\ \\ a+3d=0 \\ \\ a=-3d \\ \\ 2a=-6d \\ \\ (3-1)a=(24-30)d \\ \\ 3a-a=24d-30d \\ \\ 3a+30d=24d+a \\ \\ 3(a + 10)=a+24d \\ \\ 3 \times a_{11}=a_{25}$

$\Huge \textsc{\underline{Hence proved!!!}}$