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Question
If the pth term of an AP is q and its qth term is p then show that its (p + q)th term is zero.
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Solution
In the given AP, let the first term be a and the common difference be d.
Then Tn = a + (n - 1)d
⇒ Tp = a + (p - 1)d = q ...(i)
⇒ Tq = a + (q - 1)d = p ...(ii)
On subtracting (i) from (ii), we get:
(q - p)d = (p - q)
⇒ d = -1
Putting d = -1 in (i), we get:
a = (p + q - 1)
Thus, a = (p + q - 1) and d = -1
Now, Tp+q = a + (p + q - 1)d
= (p + q - 1) + (p + q - 1)(-1) = (p + q - 1) - (p + q - 1) = 0 Hence, the (p+q)th term is 0 (zero).
Then Tn = a + (n - 1)d
⇒ Tp = a + (p - 1)d = q ...(i)
⇒ Tq = a + (q - 1)d = p ...(ii)
On subtracting (i) from (ii), we get:
(q - p)d = (p - q)
⇒ d = -1
Putting d = -1 in (i), we get:
a = (p + q - 1)
Thus, a = (p + q - 1) and d = -1
Now, Tp+q = a + (p + q - 1)d
= (p + q - 1) + (p + q - 1)(-1)
On subtracting (i) from (ii), we get:
(q - p)d = (p - q)
⇒ d = -1
Putting d = -1 in (i), we get:
a = (p + q - 1)
Thus, a = (p + q - 1) and d = -1
Now, Tp+q = a + (p + q - 1)d
= (p + q - 1) + (p + q - 1)(-1)
= (p + q - 1) - (p + q - 1) = 0
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