Question

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 1

The given sequence is
9, 7, 5, . . .

Here,
a = 9
d = –2

We know that,
$$\begin{gathered} a_n=a+\left(n-1\right)d \\ =9+\left(n-1\right)\left(-2\right) \\ =9-2n+2 \\ =11-2n...\left(1\right) \end{gathered}$$

Another given sequence is
24, 21, 18, . . . .

Here,
a = 24
d = –3

We know that,
$$\begin{gathered} {a^{\text{'}}}_n=a+\left(n-1\right)d \\ =24+\left(n-1\right)\left(-3\right) \\ =24-3n+3 \\ =27-3n...\left(2\right) \end{gathered}$$

It is given that,
n^th term of both the progressions are equal.

From (1) and (2), we get
$$\begin{gathered} 11-2n=27-3n \\ \Rightarrow 3n-2n=27-11 \\ \Rightarrow n=16 \end{gathered}$$

Hence, n = 16.

Now,
$$\begin{gathered} a_{16}=11-2\left(16\right)\left(\mathrm{from}\left(1\right)\right) \\ =11-32 \\ =-21 \end{gathered}$$

Hence, the value of n and n^th term is 16 and –21, respectively.