Question

Given a geometric progression whose common ratio is 2, if the sum of the 6th to 7th terms is 288, determine the sum of the 4th to 15th terms inclusive.

Answers

oobleck
ar^5 + ar^6 = 288
so, with r=2,
32a+64a = 288
96a = 288
a = 3
Now, the sum from the 4th to 15th terms is
S15 - S3 = 3(2^15-1 - (2^3-1)) = _____
Bosnian
In GP n-th term:

an = a1 ∙ r ⁿ⁻¹

a1 = initial value

r = common ratio


In this case:

r = 2

a6 = a1∙ r⁶⁻¹ = a1∙ r⁵ = a1 ∙ 2⁵ = a1 ∙ 32 = 32 a1

a7 = a1 ∙ r⁷⁻¹ = a1 ∙ r⁶ = a1 ∙ 2⁶ = a1 ∙ 64 = 64 a1


a6 + a7 = 288

32 a1 + 64 a1= 288

96a1 = 288

a1 = 288 / 96

a1 = 3


a4 = a1 ∙ r⁴⁻¹ = a1 ∙ r³ = 3 ∙ 2³ = 3 ∙ 8 = 24

a15 = a1 ∙ r¹⁵⁻¹ = a1 ∙ r¹⁴ = 3 ∙ 2¹⁴ = 3 ∙ 16 384 = 49 152


a4 + a15 = 24 + 49 152 = 49 176


By the way your GP is:

3 , 6 , 12 , 24 , 48 , 96 , 192 , 384 , 768 , 1 535 , 3 072 , 6 144 , 12 288 , 24 , 576 , 49 152...


a6 + a7 = 96 + 192 = 288

a4 + a15 = 24 + 49 152 = 49 176

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