Question

A sequence can be generated by using an=a(n−1)+29, where a1=46 and n is a whole number greater than 1. What is the 6th term in the sequence, a6?

A. 180
B. 191
C. 200
D. 211

Answer 1

Given : A sequence can be generated by using aₙ=aₙ₋₁+29, where a₁=46 and n is a whole number greater than 1.  

To Find :  6th term in the sequence, a₆?

A. 180

B. 191

C. 200

D. 211

Solution:

aₙ=aₙ₋₁+29

a₆ = a₅ + 29

a₅= a₄ + 29

=> a₆ = (a₄ + 29) + 29

=> a₆ =  a₄ + 58

a₄= a₃ + 29

=>  a₆ = (a₃ + 29) + 58

=>  a₆ = a₃ +  87

a₃= a₂ + 29

=>  a₆ = a₂ + 29+  87

=>  a₆ = a₂ + 116

a₂ = a₁ + 29  => a₂ = 46 + 29  = 75

=>  a₆ =  75 + 116

=> a₆ =  191

6th term in the sequence a₆ =  191

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Answer 2

SOLUTION

TO CHOOSE THE CORRECT OPTION

A sequence can be generated by using

$\sf{a_n =a_{n - 1} + 29 \: \: \: where \: \: \: a_1 = 46}$

and n is a whole number greater than 1.

What is the 6th term in the sequence $\sf{a_6 }$

A. 180

B. 191

C. 200

D. 211

EVALUATION

Here it is given that a sequence is generated by

$\sf{a_n =a_{n - 1} + 29 \: \: \: where \: \: \: a_1 = 46}$

Thus

$\sf{a_n =a_{n - 1} + 29 }$

So we get

$\sf{a_n - a_{n - 1} = 29 }$

Putting n = 6 , 5 , 4 , 3 , 2 we get

$\sf{a_6 - a_{5} = 29 }$

$\sf{a_5 - a_{4} = 29 }$

$\sf{a_4 - a_{3} = 29 }$

$\sf{a_3 - a_{2} = 29 }$

$\sf{a_2 - a_{1} = 29 }$

Adding above equations we get

$\sf{a_6 - a_{1} = 29 \times 5 }$

$\sf{ \implies \: a_6 - a_{1} = 145 }$

$\sf{ \implies \: a_6 -46 = 145 }$

$\sf{ \implies \: a_6 = 145 + 46 }$

$\sf{ \implies \: a_6 = 191 }$

FINAL ANSWER

Hence the correct option is B. 191

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