6^{\;{th }\;} term of series a b, a² b³, a³ b^{5} is equal to

Correct Answer is : a6b11a^{6} b¹1}a6b11

Correct Answer is : a6b11a^{6} b¹1}a6b11

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Sequence and Series

Question : 13 of 86

Marks: +1, -0

6^{\;\text{th }\;}

a b, a^{2} b^{3}, a^{3} b^{5} \dots

is equal to

Solution:

Concept:

Identify the pattern of exponents for

a

and

b

in the given series.

Explanation:

The series is

a b, a^{2} b^{3}, a^{3} b^{5}, \dots

For the

n^{\text{th}}

term, the exponent of

a

n

The exponents of

b

are

1, 3, 5, \dots

, which form an arithmetic progression with first term

1

and common difference

2

Thus, the exponent of

b

for the

n^{\text{th}}

term is

1 + (n-1) \cdot 2 = 2n - 1

Hence, the

n^{\text{th}}

term is

a^{n} b^{2n-1}

For

n = 6

, the term is

a^{6} b^{2(6)-1} = a^{6} b^{11}

Answer:

a^{6}b^{11}

which corresponds to option A.

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