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.
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)) = _____
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
To find the sum of consecutive terms in a geometric progression, we can use the formula:
\[ S_n = \dfrac{a(r^n - 1)}{r - 1} \]
Where:
- \(S_n\) is the sum of the first \(n\) terms.
- \(a\) is the first term.
- \(r\) is the common ratio.
Given that the common ratio (\(r\)) is 2, we need to find the values of the 6th and 7th terms to determine the sum of the 4th to 15th terms inclusive.
Let's start by finding the 6th and 7th terms.
The formula for the \(n\)th term of a geometric progression is:
\[ T_n = a \cdot r^{n-1} \]
Plugging in the values, we have:
For the 6th term:
\[ T_6 = a \cdot (2^{6-1}) \ = 32a \]
For the 7th term:
\[ T_7 = a \cdot (2^{7-1}) \ = 64a \]
Given that the sum of the 6th and 7th terms is 288, we can write the equation:
\[ T_6 + T_7 = 32a + 64a = 288 \]
Combining like terms:
\[ 96a = 288 \]
Solving for \(a\):
\[ a = \dfrac{288}{96} = 3 \]
Now that we have \(a = 3\), we can find the values of the 4th and 15th terms and calculate the sum of the terms from the 4th to the 15th term inclusive.
The 4th term is:
\[ T_4 = a \cdot r^{4-1} = 3 \cdot 2^3 = 3 \cdot 8 = 24 \]
The 15th term is:
\[ T_{15} = a \cdot r^{15-1} = 3 \cdot 2^{14} = 3 \cdot 16384 = 49152 \]
Now, we can use the sum formula to find the sum of the 4th to 15th terms inclusive:
\[ S_{12} = \dfrac{a(r^{15} - r^3)}{r - 1} = \dfrac{3(49152 - 24)}{2 - 1} = \dfrac{3 \cdot 49128}{1} = 147384 \]
Therefore, the sum of the 4th to 15th terms inclusive is 147384.
To find the sum of terms in a geometric progression, we can use the formula:
S = a * (r^n - 1) / (r - 1)
where S is the sum of terms, a is the first term, r is the common ratio, and n is the number of terms.
In this case, the common ratio (r) is 2. We need to find the sum of the 4th to 15th terms inclusive.
Step 1: Find the value of the 6th term.
Since the common ratio is 2, we can calculate:
a * (r^(6 - 1)) = 288
Substituting r = 2:
a * 2^5 = 288
a * 32 = 288
Dividing both sides by 32:
a = 9
Step 2: Calculate the sum of the 4th to 15th terms.
Using the formula for the sum of terms in a geometric progression, we have:
S = a * (r^n - 1) / (r - 1)
Substituting a = 9, r = 2, and n = 15:
S = 9 * (2^15 - 1) / (2 - 1)
Simplifying:
S = 9 * (32768 - 1)
S = 9 * 32767
S = 294903
Therefore, the sum of the 4th to 15th terms inclusive is 294903.