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Answer :
The sum of the sixth terms of the GP is 124. The common difference is 3 and the first term is 4.
To find the sum of the sixth terms of the geometric progression (GP), we can use the formula for the sum of the first n terms of a GP: [tex]S_n = a(r^n - 1) / (r - 1)[/tex].
Plugging in the values a = 4, r = 2, and n = 6, we get: [tex]S_6 = 4(2^6 - 1) / (2 - 1) = 124[/tex]
Therefore, the sum of the sixth terms of the GP is 124.
To find the common difference (d) and first term (a) of the arithmetic progression (AP), we can use the formula for the sum of the first n terms of an AP:
[tex]Sn = (n/2)(2a + (n-1)d)[/tex]
Given that [tex]S_{10}[/tex] = 175, we can solve for d and a.
Plugging in n = 10 and [tex]S_{10}[/tex] = 175, we get:
175 = (10/2)(2a + 9d).
Simplifying, we get:
175 = 5(2a + 9d).
Now we need to use the information that the third term of the GP is the fifth term of the AP. The third term of the GP is [tex]ar^2[/tex], which is equal to the fifth term of the AP. So we can write the equation:
[tex]ar^2[/tex] = a + 4d
Substituting a = (175 - 45d)/10 into this equation, we can solve for d.
[tex]\frac{(175 - 45d)}{10} r^2 = \frac{(175 - 45d)}{10} + 4d[/tex]
[tex]\frac{(175 - 45d)}{10} 4 = \frac{(175 - 45d)}{10} + 4d[/tex]
[tex]\frac{(175 - 45d)}{10} (4-1) = 4d[/tex]
[tex]\frac{(175 - 45d)}{10} 3 = 4d[/tex]
[tex](175 - 45d) = \frac{4d \times 10}{3}[/tex]
[tex]175 = 13.3d + 45d[/tex]
[tex]58.3d = 175[/tex]
d =3
The first term of AP:
a = (175 - 45d)/10
a = (175 - 45 x 3)/10
a = (175 - 135)/10
a = 40/10
a = 4
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(a) The sum of the sixth terms of GP is 252
(b) The common difference and the first term of the AP are 3 and 4 respectively
Step-by-step explanation:
The formula of the nth term of an Geometric Progression (GP) is
[tex]a_{n}=ar^{n-1}[/tex] , where
- a is the first term
- r is the common ratio between each two consecutive terms
The formula of the sum of n terms of an Geometric Progression is
[tex]S_{n}=\frac{a(r^{n}-1)}{r-1}[/tex]
The formula of the nth term of an Arithmetic Progression (AP) is
[tex]a_{n}=a+(n-1)d[/tex] , where
- a is the first term
- d is the common difference between each two consecutive terms
The formula of the sum of n terms of an Arithmetic Progression is
[tex]S_{n}=\frac{n}{2}[2a+(n-1)d][/tex]
∵ A GP has the first term and common ratio 4 and 2 respectively
∴ a = 4 and r = 2
- Find the third term of it
∵ [tex]a_{n}=ar^{n-1}[/tex]
∵ n = 3
- Substitute the values of a, r, n in the formula to find the 3rd term
∴ [tex]a_{3}=4(2)^{3-1}=4(2)^{2}=4(4)=16[/tex]
∴ The third term is 16
∵ The sum of n terms in GP is [tex]S_{n}=\frac{a(r^{n}-1)}{r-1}[/tex]
∵ n = 6 , a = 4 and r = 2
- Substitute the values of a, r, n in the formula to find [tex]S_{6}[/tex]
∴ [tex]S_{6}=\frac{4[(2)^{6}-1]}{2-1}[/tex]
∴ [tex]S_{6}=\frac{4[64-1]}{1}[/tex]
∴ [tex]S_{6}=4[63][/tex]
∴ [tex]S_{6}=252[/tex]
(a) The sum of the sixth terms of GP is 252
∵ The third term of the GP is the fifth term of an AP
∵ The third term of the GP = 16
∴ The 5th term of AP = 16
∵ [tex]a_{n}=a+(n-1)d[/tex]
∵ n = 5
∵ [tex]a_{5}=a+(5-1)d[/tex]
∴ [tex]a_{5}=a+4d[/tex]
∵ [tex]a_{5}=16[/tex]
- Equate the two expressions of [tex]a_{5}[/tex]
∴ a + 4d = 16 ⇒ (1)
∵ The sum of the first ten terms of the AP is 175
∴ n = 10
∵ [tex]S_{10}=\frac{10}{2}[2a+(10-1)d][/tex]
∴ [tex]S_{10}=5[2a+9d][/tex]
∵ [tex]S_{10}=175[/tex]
- Equate the two expressions of [tex]S_{10}[/tex]
∴ 5[2a + 9d] = 175
- Divide both sides by 5 to simplify the equation
∴ 2a + 9d = 35 ⇒ (2)
Now we have a system of equation to solve it
Multiply equation (1) by -2 to eliminate a
∵ -2a - 8d = -32 ⇒ (3)
- Add equations (2) and (3)
∴ d = 3
- Substitute the value of d in equation (1) or (2) to find a
∵ a + 4(3) = 16
∴ a + 12 = 16
- Subtract 12 from both sides
∴ a = 4
(b) The common difference and the first term of the AP are 3 and 4 respectively
Learn more:
You can learn more about the AP and GP in brainly.com/question/7221312
brainly.com/question/1522572
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