Question

The 4th team of AP is zero .prove that the 25th term of AP is three times ita 11th term

Answer 1

$\underline{\underline{\mathfrak{\Large{Solution : }}}}$


$\textsf{Let , first term is a and common difference} \\ \textsf{is d.}$



$\underline{\textsf{Given,}} \\ \\ \sf \implies T_4 \: = \: 0 \\ \\ \sf \implies a \: + \: ( \: 4 \: - \: 1 )d \: = \: 0 \\ \\ \sf \implies a \: + \: 3d \: = \: 0 \\ \\ \sf \: \: \therefore \: \: a \: = \: -3d \qquad...(1)$



$\underline{\mathsf{To \: Prove \: \longrightarrow T_{25} \: = \: 3(T_{11})}}$



$\underline{\textsf{Now,}} \\ \\ \sf \implies T_{25} \: = \: a \: + \: (25 \: - \: 1)d \\ \\ \sf \implies T_{25} \: = \: a \: + \: 24d \\ \\ \textsf{Plug the value of (1),} \\ \\ \sf \implies T_{25} \: = \: - \: 3d \: + \: 24d \\ \\ \: \: \sf \therefore \: \: T_{25} \: = \: -21d \quad...(2)$



$\underline{\textsf{Again,}} \\ \\ \sf \implies T_{11} \: = \: a \: + \: (11\: - \: 1)d \\ \\ \sf \implies T_{11} \: = \: a \: + \: 10d \\ \\ \textsf{Plug the value of (1),} \\ \\ \sf \implies T_{11} \: = \: - \: 3d \: + \: 20d \\ \\ \: \: \sf \therefore \: \: T_{11} \: = \: -7d \quad...(3)$





$\underline{\textsf{Divide (2) by (3), }} \\ \\ \sf \implies \dfrac{T_{25}}{T_{11}} \: = \: \dfrac{-21d}{-7d} \\ \\ \sf \implies \dfrac{T_{25}}{T_{11}} \: = \: 3 \\ \\ \sf \: \: \therefore\: \: T_{25} \: = \: 3(T_{11}) \\ \\ \\ \underline{\underline{\mathsf{\Large{Proved !! }}} }$






$\underline{\textsf{Formulae related to A.P.}} \\ \\ \sf \implies T_n \: = \: a \: + \: (n \: - \: 1)d \\ \\ \sf \implies T_n \: = \: S_n \: - \: S_{(n \: - \: 1)} \\ \\ \sf \implies d \: = \: T_{n} \: - \: T_{(n \: - \: 1)} \\ \\ \sf \implies S_n \: = \: \dfrac{n}{2}\{2a \: + \: (n \: - \: 1)d \} \\ \\ \sf \qquad \qquad \qquad or \\ \\ \sf \qquad \qquad = \dfrac{n}{2}(a \: + \: l)$