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If {(p+1)}^{th} term of an A.P is twice {(q+1)}^{th} term, Pro

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If $(p + 1)^{t h}$ term of an A.P is twice $(q + 1)^{t h}$ term, Prove that $(3 p + 1)^{t h}$ term is twice the $(p + q + 1)^{t h}$ term.

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Text solutionVerified

Let the first term of the A.P be

a

and the common difference be

d

We have

t_{p + 1} = 2 t_{q + 1}

\Rightarrow a + (p + 1 - 1) d = 2 [a + (q + 1 - 1) d]

\Rightarrow a + p d = 2 a + 2 q d

\Rightarrow a = (p - 2 q) d

..........

(1)

Now,

t_{3 p + 1} = a + (3 p + 1 - 1) d

= (p - 2 q) d + 3 p d = 4 p d - 2 q d = 2 (2 p - q) d

∴ t_{3 p + 1} = 2 (2 p - q) d

......

(2)

and

t_{p + q + 1} = a + (p + q + 1 - 1) d = (p - 2 q) d + (p + q) d

from

(1)

\Rightarrow t_{p + q + 1} = (2 p - q) d

.....

(3)

From

(2)

and

(3)

we have

t_{3 p + 1} = 2 t_{p + q + 1}

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Question Text	If $(p + 1)^{t h}$ term of an A.P is twice $(q + 1)^{t h}$ term, Prove that $(3 p + 1)^{t h}$ term is twice the $(p + q + 1)^{t h}$ term.
Answer Type	Text solution:1
Upvotes	150