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Mathematics

The 19th term of an A.P. is equal to three times its 6th term. If its 9th term is 19, find the A.P.

AP GP

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Answer

We know that

    an = a + (n - 1)d

According to question,

⇒ a19 = 3(a6)
⇒ a + 18d = 3(a + 5d)
⇒ a + 18d = 3a + 15d
⇒ 3a - a = 18d - 15d
⇒ 2a = 3d
⇒ a = 32d\dfrac{3}{2}d     (Eq 1)

Given, a9 = 19

a+8d=1932d+8d=193d+16d2=1919d2=19d=1919×2d=2.\Rightarrow a + 8d = 19 \\[1em] \Rightarrow \dfrac{3}{2}d + 8d = 19 \\[1em] \Rightarrow \dfrac{3d + 16d}{2} = 19 \\[1em] \Rightarrow \dfrac{19d}{2} = 19 \\[1em] \Rightarrow d = \dfrac{19}{19} \times 2 \\[1em] \Rightarrow d = 2. \\[1em]

Putting value of d in Eq 1 we get,

a=32da=32×2a=3.\Rightarrow a = \dfrac{3}{2}d \\[1em] \Rightarrow a = \dfrac{3}{2} \times 2 \\[1em] \Rightarrow a = 3.

Hence, the terms of A.P. are

a2 = a1 + d = 3 + 2 = 5,
a3 = a2 + d = 5 + 2 = 7,
a4 = a3 + d = 7 + 2 = 9.

Hence, the terms of the A.P. are 3, 5, 7, 9, ….

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