The 19th term of an A.P. is equal to three times its sixth term.

Question:

The $19^{\text {th }}$ term of an A.P. is equal to three times its sixth term. If its $9^{\text {th }}$ term is 19 , find the A.P.

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,

a19 = 3a6
⇒ a + (19 − 1)d =  3(a + (6 − 1)d)
⇒ a + 18d =  3a + 15d
⇒ 18d − 15=  3a − a
⇒ 3= 2a
⇒ a = 32$=\frac{3}{2} d$  .... (1)

Also, a9 = 19
⇒ a + (9 − 1)d = 19
⇒ a + 8d = 19   ....(2)

On substituting the values of (1) in (2), we get
32$\frac{3}{2} d$ + 8d = 19
⇒ 3+ 16= 19 × 2
⇒ 19= 38
⇒ = 2
⇒ a = 32×2$=\frac{3}{2}$×2     [From (1)]
⇒ a = 3

Thus, the A.P. is 3, 5, 7, 9, .... .

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