If an is a sequence with a0-p and an-an-1-ran-1 for n- 1- then the terms of the sequence are in

Explanation for correct option:Step 1: Simplify the given data.Given, a_0=p &a_n a_n 1=ra_n 1⇒a_n=ra_n 1+a_n 1⇒a_n=a_n 1(r+1)Step 2: Substitute the value of n=1 ...

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Standard XII

Mathematics

nth Term of A.P

If an is a se...

Question

If $\{a_{n}\}$ is a sequence with $a_{0} = p$ and $a_{n} - a_{(n - 1)} = r a_{(n - 1)}$ for $n \geq 1$ , then the terms of the sequence are in

A.P

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G.P

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H.P

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an Arithmetic – geometric progression

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Solution

The correct option is B
G.P

Explanation for correct option:
Step-1: Simplify the given data.
Given, $a_{0} = p & a_{n} - a_{n - 1} = r a_{n - 1}$
$\Rightarrow$ $a_{n} = r a_{n - 1} + a_{n - 1}$
$\Rightarrow$ $a_{n} = a_{n - 1} (r + 1)$
Step-2: Substitute the value of $n = 1, 2, 3$ ,
$a_{1} = a_{0} (r + 1) = p (r + 1)$
$a_{2} = a_{1} (r + 1) = p {(r + 1)}^{2}$
$a_{3} = a_{2} (r + 1) = p {(r + 1)}^{3}$
Step-3: Check Relation between $a_{1}, a_{2}, a_{3}$ .
$\Rightarrow$ $\frac{p {(r + 1)}^{2}}{p (r + 1)} = \frac{p {(r + 1)}^{3}}{p {(r + 1)}^{2}}$
$\Rightarrow$ $(r + 1) = (r + 1)$
$\Rightarrow$ $\frac{a_{2}}{a_{1}} = \frac{a_{3}}{a_{2}} = r$
The terms of the sequence are in GP.
Hence, correct answer is option $(B)$ .

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If an is a sequence with a0=p and an-an-1=ran-1 for n≥1, then the terms of the sequence are in

If $\{a_{n}\}$ is a sequence with $a_{0} = p$ and $a_{n} - a_{(n - 1)} = r a_{(n - 1)}$ for $n \geq 1$ , then the terms of the sequence are in