Asked by Jane
                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
                    Answered by
            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)) = _____
    
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)) = _____
                    Answered by
            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
    
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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