What is the sum to 5 terms of the geometric progression whose first term is 54 and fourth term is 2?
same as before.
54(1 - 1/3^4)/(1 - 1/3)
a1=54 ,a4= 2
a1 + 3d =2
a1 =54
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3d = -52 d = -52/3
s5 =n/2[2a+(n-1)d] = 5/2[2x54+ 4x-52/3]
= 5/2[108-208/3] = 5/2[324-208/3] =5/2[116/3] =5x116/2x3
= 580/6 = 290/3
To find the sum of the first 5 terms of a geometric progression, we need the first term (a) and the common ratio (r).
Given that the first term (a) is 54 and the fourth term is 2, we can use this information to determine the common ratio (r).
The fourth term (a₄) can be expressed as:
a₄ = a * r³
Substituting the given values, we have:
2 = 54 * r³
To find the value of r, we can rearrange the equation to isolate r:
r³ = 2/54
r³ = 1/27
Taking the cube root of both sides, we get:
r = ∛(1/27)
r = 1/3
Now that we have determined the value of r, we can find the sum of the first 5 terms of the geometric progression using the formula:
S₅ = a * (r⁵ - 1) / (r - 1)
Substituting the given values, we have:
S₅ = 54 * ((1/3)⁵ - 1) / ((1/3) - 1)
Calculating the terms within the parentheses first:
(1/3)⁵ = 1/243
Now substituting this value and simplifying further:
S₅ = 54 * (1/243 - 1) / (-2/3)
S₅ = 54 * (1/243 - 243/243) / (-2/3)
S₅ = 54 * (-242/243) / (-2/3)
S₅ = 54 * (-242/243) * (-3/2)
S₅ = 54 * (242/243) * (3/2)
S₅ = (13122/81) * (3/2)
S₅ = 19683/54
Therefore, the sum of the first 5 terms of the given geometric progression is 19683/54.