A sequence $a_{1}, a_{2}, a_{3}, . . . . . . .$ is defined by letting $a_{1} = 3$ and $a_{k} = 7$ $a_{k - 1}$ for all natural numbers $k \geq 2$ . Show that $a_{n} = {3.7}^{n - 1}$ for all $n ϵ N$ .

Open in App

Solution

A sequence $a_{1}, a_{2}, a_{3}, . . . .$ is defined by letting $a_{1} = 3$ and $a_{k} = 7 a_{k - 1}$ , for all natural numbers $k \geq 2$ .

Let P(n) : $a_{n} = {3.7}^{n - 1}$ for all naturel numbers.

Step I We observe P(2) is true.

For n = 2,

$a_{2} = {3.7}^{2 - 1} = {3.7}^{1} = 21$ is true.

As $a_{1} = 3. a_{k} = 3.7 a_{k - 1}$

$\Rightarrow a_{2} = 7. a_{2 - 1} = 7. a_{1}$

$\Rightarrow a_{2} = 7 \times 3 = 21 [∵ a_{1} = 3]$

Step II Now, assume that P(n) is true for n = k

P(k) : $a_{k} = 3.7 k^{- 1}$

Step III Now, to prove $P (k + 1)$ is true, we have to show that

$P (k + 1) : a_{k} = {3.7}^{k + 1 - 1}$

$a_{k + 1} = 7. a_{k + 1 - 1} = 7. a_{k}$

$= {7.3.7}^{k - 1} = {3.7}^{k - 1 + 1}$

So, P(k + 1) is true, whenever p(k) is true.

Hence, P(n) is true.

Suggest Corrections

Similar questions

A sequence $x_{1}, x_{2}, x_{3}, . . . . . .$ is defined by letting $x_{1} = 2$ and $x_{k} = \frac{x_{k - 1}}{n}$ for all natural numbers k, $k \geq 2$ . Show that $x_{n} = \frac{2}{n!}$ for all $n ϵ N$ .

Let a sequence be defined by

a_{1} = 0

and

a_{n + 1} = a_{n} + 4 n + 3

for all

n \geq 1 (n ϵ N)

The value of

a_{k}

in terms of k is

(k \in N)

A sequence x_{1}, x_{2}, x_{3}, . . . is defined by letting x_{1} = 2 and x_{k} = \frac{x_{k - 1}}{k} for all natural numbers k, k \geq 2 . Show that x_{n} = \frac{2}{n!} for all n \in N .

[NCERT EXEMPLAR]

Q. Let a sequence be defined by

a_{1} = 1, a_{2} = 1

and,

a_{n} = a_{n - 1} + a_{n - 2}

for all

n > 2

,
find

\frac{a_{n + 1}}{a_{n}}

for

n = 1, 2, 3, 4

Q. If

a_{1}, a_{2}, a_{3}, . . ., a_{n}

are in A.P. and

a_{i} > 0

for all i, then show that

\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + . . . + \frac{1}{a_{n - 1} a_{n}} = \frac{n - 1}{a_{1} a_{n}}

Join BYJU'S Learning Program

A sequence a1,a2,a3,....... is defined by letting a1=3 and ak=7 ak−1 for all natural numbers k≥2. Show that an=3.7n−1 for all nϵN.

A sequence $a_{1}, a_{2}, a_{3}, . . . . . . .$ is defined by letting $a_{1} = 3$ and $a_{k} = 7$ $a_{k - 1}$ for all natural numbers $k \geq 2$ . Show that $a_{n} = {3.7}^{n - 1}$ for all $n ϵ N$ .