If the 9th term of an AP is zero, then prove that its 29th term is twice its 19th term.

Explanation

Let the first term, common difference and number of terms of an AP are a, d and n, respectively.
Given that, 9th term of an AP,`T_(9)=0 " " [:' n"th term of an AP", T_(n)=a+(n-1)d]`
`implies a+(9-1)d=0`
`impliesa+8d=0 implies a= -8d " " …(i)`
Now, its ` " " `19th term, `T_(19)=a+(19-1)d`
` " " = -8d+18d " " ` [from Eq. (i)]
` " " =10d " " `...(ii)
and its ` " " `29th term, `T_(29)=a+(29-1)d`
` " " = -8d+28d " " ` [from Eq. (i)]
` " " =20d=2xx(10d)`
`implies " " T_(29)=2xxT_(19)`
Hence, its 29th term is twice its 19th term. ` " Hence proved." `