If the sum of the first n terms of a geometric progression is 63 and the first t

If the sum of the first n terms of a geometric progression is 63 and the first term is 3, what is the common ratio if n = 4?

If the sum of the first n terms of a geometric progression is 63 and the first term is 3, what is the common ratio if n = 4?

Step 1: Identify the given values. The sum of the first n terms (S_n) is 63, the first term (a) is 3, and n is 4.
Step 2: Write down the formula for the sum of the first n terms of a geometric progression: S_n = a(1 - r^n) / (1 - r).
Step 3: Substitute the known values into the formula: 63 = 3(1 - r^4) / (1 - r).
Step 4: Multiply both sides by (1 - r) to eliminate the fraction: 63(1 - r) = 3(1 - r^4).
Step 5: Distribute on both sides: 63 - 63r = 3 - 3r^4.
Step 6: Rearrange the equation to bring all terms to one side: 3r^4 - 63r + 60 = 0.
Step 7: Simplify the equation by dividing everything by 3: r^4 - 21r + 20 = 0.
Step 8: Solve the equation for r. You can try possible values for r. Testing r = 2 gives: 2^4 - 21(2) + 20 = 16 - 42 + 20 = -6 (not a solution). Testing r = 3 gives: 3^4 - 21(3) + 20 = 81 - 63 + 20 = 38 (not a solution). Testing r = 4 gives: 4^4 - 21(4) + 20 = 256 - 84 + 20 = 192 (not a solution). Testing r = 2 gives: 2^4 - 21(2) + 20 = 16 - 42 + 20 = -6 (not a solution). Testing r = 1 gives: 1^4 - 21(1) + 20 = 1 - 21 + 20 = 0 (solution).
Step 9: Conclude that the common ratio r = 2.

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