Question

In a geometric series 2+4+8+...........,starting from the first term how many consecutive terms are needed to yield the sum 1022?

Answer 1

Answer:

Let, n consecutive terms are needed to yield the sum of 1022.

Then, we have,

2+4+8+….upto n terms=10222+4+8+….upto n terms=1022

⟹2⋅(2n−1)2–1=1022⟹2⋅(2n−1)2–1=1022

[As, the first term and common ratio of the GP are 2 and 2 respectively.]

⟹2n−1=511⟹2n−1=511

⟹2n=512=29⟹2n=512=29

∴n=9.†∴n=9.†

Hence, 9 consecutive terms are needed to yield the sum of 1022.

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