Two A.P's are given 9,7,5..and 24,21,18,….If `n^(th)`term of both the progressions are equal then find the value of n and `n^(th)` term

Answer

The value of n is 16 and the 16th term is -21.

Explanation

The given A.P. is 9, 7, 5, …
Here, `a=9, d=7 -9=-2`
`t_(n)=a=(n-1)d` …(Formula)
`=9+(n-1)xx(-2)`
`=9-2n+2`
`thereforet_(n)=11-2n" "...(1)`
The second A.P. is `24,21,18….`
Here, `a=24, d=21-24=-3`
`t_(n)=a+(n-1)d` ...(Formula)
`=24+(n-1)xx(-3)`
`=24-3n+3`
`thereforet_(n)=27-3n" "...(2)`
From the given condition,
`11-2_(n)=27-3_(n)`
`therefore-2_(n)+3_(n)=27-11`
`thereforen=16`
Now, we find `t_(16)`.
`t_(16)=9+(16-1)xx(-2)`
`=9+15xx(-2)`
`=9-30`
`thereforet_(16)=-21`