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Question
Two A.P.’ s are given 9, 7, 5, . . . and 24, 21, 18, . . . . If nth term of both the progressions are equal then find the value of n and nth term.
Solution
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 nth term of both the progressions is equal.
From (1) and (2), we get
\[11 - 2n = 27 - 3n\]
\[ \Rightarrow 3n - 2n = 27 - 11\]
\[ \Rightarrow n = 16\]
nth term = an = 16th term
Hence, n = 16.
Now,
an = a + (n - 1)d
a16 = 9 + (16 - 1) × (-2)
a16 = 9 + (15) × (-2)
a16 = 9 - 30
a16 = -21
Hence, the value of n = 16 and the nth term a16 = –21.
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