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Question
The 9th term of an A.P. is equal to 6 times its second term. If its 5th term is 22, find the A.P.
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Solution
Let a be the first term and d be the common difference.
We know that, nth term = an = a + (n − 1)d
According to the question,
a9 = 6a2
⇒ a + (9 − 1)d = 6(a + (2 − 1)d)
⇒ a + 8d = 6a + 6d
⇒ 8d − 6d = 6a − a
⇒ 2d = 5a
⇒ a = \[\frac{2}{5}\]d .... (1)
Also, a5 = 22
⇒ a + (5 − 1)d = 22
⇒ a + 4d = 22 ....(2)
On substituting the values of (1) in (2), we get \[\frac{2}{5}\]d + 4d = 22
⇒ 2d + 20d = 22 × 5
⇒ 22d = 110
⇒ d = 5
⇒ a =\[\frac{2}{5} \times 5\] [From (1)]
⇒ a = 2
Thus, the A.P. is 2, 7, 12, 17, .... .
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