In a sequence of numbers the first number is $3$ and each number after the first is $2$ more than $3$ times the preceding number. What is the fourth term in the sequence?

105

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106

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107

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108

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Solution

The correct option is D $107$
First number $= 3$
According to the question,
Second number $= 3 \times 3 + 2 = 9 + 2 = 11$
$∴$ for any number $n, t_{n} = 3 \times t_{n - 1} + 2$
So, third number $=$ $11 \times 3 + 2 = 33 + 2 = 35$
And, fourth number $=$ $35 \times 3 + 2 = 105 + 2 = 107$
$∴$ fourth term in the sequence $= 107$

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In a sequence of numbers the first number is 3 and each number after the first is 2 more than 3 times the preceding number. What is the fourth term in the sequence?

The correct option is D 107First number =3According to the question, Second number =3×3+2=9+2=11∴ for any number n,tn=3×tn−1+2So, third number = 11×3+2=33+2=35And, fourth number = 35×3+2=105+2=107∴ fourth term in the sequence =107

In a sequence of numbers the first number is $3$ and each number after the first is $2$ more than $3$ times the preceding number. What is the fourth term in the sequence?