What is the sum of the first 6 terms of the geometric progression 1, 4, 16, 64, …?
Answer and explanation
Correct answer: 1365
The governing idea is the finite sum of a geometric progression. In this sequence, the first term is a = 1 and the common ratio is r = 4. For r ≠ 1, the sum of the first n terms is Sₙ = a(rⁿ − 1)/(r − 1). Hence S₆ = 1(4⁶ − 1)/(4 − 1) = (4096 − 1)/3 = 4095/3 = 1365. A direct check gives 1 + 4 + 16 + 64 + 256 + 1024 = 1365. Thus option A is correct. The nearby values in options B, C, and D result from an arithmetic error, such as failing to subtract 1 or dividing incorrectly by 3.
Frequently asked questions
What is the correct answer to this question?
1365
Why is this the correct answer?
The governing idea is the finite sum of a geometric progression. In this sequence, the first term is a = 1 and the common ratio is r = 4. For r ≠ 1, the sum of the first n terms is Sₙ = a(rⁿ − 1)/(r − 1). Hence S₆ = 1(4⁶ − 1)/(4 − 1) = (4096 − 1)/3 = 4095/3 = 1365. A direct check gives 1 + 4 + 16 + 64 + 256 + 1024 = 1365. Thus option A is correct. The nearby values in options B, C, and D result from an arithmetic error, such as failing to subtract 1 or dividing incorrectly by 3.
Which subject and chapter does this question cover?
This is a Class 9 Mathematics question. Chapter: Sequences and Progressions. Topic: Geometric Progression.
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