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In a geometric progression, the sum of the first 5 term is 31/3 and the sum of t...
Aug 5, 2024
In a geometric progression, the sum of the first 5 term is 31/3 and the sum of the next 5 terms is 992/3. Find the first term.
Solution by Steps
step 1
Given the sum of the first 5 terms of a geometric progression is 313\frac{31}{3} and the sum of the next 5 terms is 9923\frac{992}{3}. We need to solve the equations: 313=a1−r51−r \frac{31}{3} = a \frac{1 - r^5}{1 - r} 9923=ar51−r51−r \frac{992}{3} = a r^5 \frac{1 - r^5}{1 - r}
step 2
From the first equation: 313=a1−r51−r \frac{31}{3} = a \frac{1 - r^5}{1 - r} Solving for aa: a=313⋅1−r1−r5 a = \frac{31}{3} \cdot \frac{1 - r}{1 - r^5}
step 3
Substitute aa into the second equation: 9923=(313⋅1−r1−r5)r51−r51−r \frac{992}{3} = \left( \frac{31}{3} \cdot \frac{1 - r}{1 - r^5} \right) r^5 \frac{1 - r^5}{1 - r} Simplify: 9923=313r5 \frac{992}{3} = \frac{31}{3} r^5 992=31r5 992 = 31 r^5 r5=99231 r^5 = \frac{992}{31} r5=32 r^5 = 32 r=2 r = 2
step 4
Substitute r=2r = 2 back into the equation for aa: a=313⋅1−21−25 a = \frac{31}{3} \cdot \frac{1 - 2}{1 - 2^5} a=313⋅−11−32 a = \frac{31}{3} \cdot \frac{-1}{1 - 32} a=313⋅−1−31 a = \frac{31}{3} \cdot \frac{-1}{-31} a=313⋅131 a = \frac{31}{3} \cdot \frac{1}{31} a=13 a = \frac{1}{3}
Answer
The first term aa is 13\frac{1}{3}
Key Concept
Geometric Progression Sum Formula
Explanation
The sum of the first nn terms of a geometric progression can be found using the formula Sn=a1−rn1−rS_n = a \frac{1 - r^n}{1 - r}, where aa is the first term and rr is the common ratio.
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