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प्रश्न
The 19th term of an A.P. is equal to three times its sixth term. If its 9th term is 19, find the A.P.
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उत्तर
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,
a19 = 3a6
⇒ a + (19 − 1)d = 3(a + (6 − 1)d)
⇒ a + 18d = 3a + 15d
⇒ 18d − 15d = 3a − a
⇒ 3d = 2a
⇒ a = \[\frac{3}{2}\] d .... (1)
Also, a9 = 19
⇒ a + (9 − 1)d = 19
⇒ a + 8d = 19 ....(2)
On substituting the values of (1) in (2), we get
⇒ 3d + 16d = 19 × 2
⇒ 19d = 38
⇒ d = 2
⇒ a = \[\frac{3}{2} \times 2\] [From (1)]
⇒ a = 3
Thus, the A.P. is 3, 5, 7, 9, .......
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Q.2
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