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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.
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Solution
The given sequence is 9, 7, 5, ...
Here, a = 9, d = 7 – 9 = –2
tn = a + (n – 1)d ...(Formula)
= 9 + (n – 1) × (–2)
= 9 – 2n + 2
∴ tn = 11 – 2n ...(1)
The second A.P. is 24, 21, 18, ...
Here, a = 24, d = 21 – 24 = –3
tn = a + (n – 1)d ...(Formula)
= 24 + (n – 1) × (–3)
= 24 – 3n + 3
∴ tn = 27 – 3n ...(2)
From the given condition,
11 – 2n = 27 – 3n
∴ –2n + 3n = 27 – 3n
∴ n = 16
Now we find t16
tn = a + (n – 1)d ...(Formula)
∴ t16 = 9 + (16 – 1) × (–2) ...(Taking values from 1st A.P.)
= 9 + 15 × (–2)
= 9 – 30
∴ t16 = –21
∴ The value of n is 16, and the nth (16th) term is –21.
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