If the `n`th terms of the two AP's 9, 7, 5, … and 24, 21, 18, … are the same, then find the value of `n`. Also, that term.

Explanation

Let the first term, common difference and number of terms of the AP 9, 7, 5, … are `a_(1), d_(1),` and `n_(1),` respectively.
i.e., first term `(a_(1))=9` and common difference `(d_(1))=7-9= -2.`
` :. ` Its `n`th term, ` T_(n_(1))^("'")=a_(1)+(n_(1)-1)d_(1)`
`impliesT_(n_(1))^("'")=9+(n_(1)-1)(-2)`
`impliesT_(n_(1))^("'")=9-2n_(1)+2`
`impliesT_(n_(1))^("'")=11-2n_(1) " " `[` :' n`th term of an AP, `T_(n)=a+(n-1)d`] ...(i)
Let the first term, common difference and the number of terms of the AP 24, 21, 18, ... are `a_(2), d_(2)` and `n_(2),` respectively.
i.e., first term `(a_(2))=24` and common difference `(d_(2))=21-24= -3.`
` :. ` Its `n`th term, ` T_(n_(2))^("''")=a_(2)+(n_(2)-1)d_(2)`
`impliesT_(n_(2))^("''")=24+(n_(2)-1)(-3)`
`impliesT_(n_(2))^("''")=24-3n_(2)+3`
`impliesT_(n_(2))^("''")=27-3n_(2) " " ` ...(ii)
Now by given condition,
`n`th terms of the both APs are same, i.e., ` T_(n_(1))^("'")=T_(n_(2))^("''")`
`11-2n_(1)=27-3n_(2) " " ` [from Eqs. (i) and (ii)]
`implies n=16`
` :. n`th term of first AP,` T_(n_(1))^("'")=11-2n_(1)=11-2(16)=11-32= -21`
and `n`th term of second AP,` T_(n_(2))^("''")=27-3n_(2)=27-3(16)=27-48= -21`
Hence, the value of `n` is 16 and that term i.e., `n`th term is -21 .