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,

\[a_n = a + \left( n - 1 \right)d\]

\[ = 9 + \left( n - 1 \right)\left( - 2 \right)\]

\[ = 9 - 2n + 2\]

\[a_n\] \[ = 11 - 2n . . . \left( 1 \right)\]

Another given sequence is
24, 21, 18, . . . .
Here,
a = 24
d = 21 - 24 =  –3
We know that,

\[a_n = a + \left( n - 1 \right)d\]

\[ = 24 + \left( n - 1 \right)\left( - 3 \right)\]

\[ = 24 - 3n + 3\]

\[a_n\] \[ = 27 - 3n . . . \left( 2 \right)\]

It is given that,
the n^th term of both the progressions is equal.
From (1) and (2), we get

\[11 - 2n = 27 - 3n\]

\[ \Rightarrow 3n - 2n = 27 - 11\]

\[ \Rightarrow n = 16\]

n^th term = a_n = 16^th term

Hence, n = 16.

Now,

a_n = a + (n - 1)d

a₁₆ = 9 + (16 - 1) × (-2)

a₁₆ = 9 + (15) × (-2)

a_{16 =} 9 - 30

a₁₆ = -21

Hence, the value of n = 16 and the n^th term a₁₆ = –21.