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.

Answer 1

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)) = _____

Answer 2

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