The GMAT Math practice question given below is a sequences and series question. Concept covered is sum of a Geometric Progression and common ratio of a GP and working with algebraic identities. A medium difficulty GMAT 625 level problem solving question.

Question 20: If the ratio of the sum of the first 6 terms of a GP to the sum of the first 3 terms of the GP is 9, what is the common ratio of the Geometric Progression?

- 3
- \\frac {1} {3})
- 2
- 9
- \\frac {1} {9})

@ INR

Formula to find the sum of first 'n' terms of a GP

The sum of the first n terms of a G.P. is given by \\frac {{{ar}^{n}}-a} {r-1}), where 'a' is the first term of the G.P., 'r' is the common ratio and 'n' is the number of terms in the G.P.

Therefore, the sum of the first 6 terms of the G.P will be equal to \\frac {a\left({r}^{6}-1\right)} {r-1})

And sum of the first 3 terms of the G.P. will be equal to \\frac {a\left ( r^3-1\right)} {r-1})

Use the ratio between these two sums to find 'r'

The ratio of the sum of the first 6 terms : sum of first 3 terms = 9 : 1

i.e., \\frac {\frac {a\left(r^6 - 1\right)} {r - 1}} {\frac {a\left(r^3 - 1\right)} {r - 1}}) = 9

\\frac {{{r}^{6}}-1} {{{r}^{3}}-1}) = \\frac {\left ( {{{r}^{3}}+1} \right )\left ( {{{r}^{3}}-1} \right )} {{{r}^{3}}-1}) = 9

Or r^{3} + 1 = 9

r^{3} = 8

r = 2

Many of us know the formula and may have applied the sum of GP formula many a times. However, we may not have known how the sum of GP formula was derived. Understanding the derivation of the formula for the sum of a finite geometric progression (G.P.) provides valuable insight into how sequences in mathematics operate. Let's break down the formula derivation step by step for clarity and comprehension.

Consider a geometric sequence where:

a is the first term

r is the common ratio, and

n is the number of terms.

The sequence looks like this: (a, ar, ar^{2}, ar^{3}, ...... ar^{(n - 1)}

**Step 1: Write the Sum Equation**

The sum of the first n terms S_{n} can be expressed as: S_{n} = a + ar + ar^{2} + ar^{3} + ..... + ar^{(n - 1)} ---- (1)

**Step 2: Multiply both sides of equation (1) by the Common Ratio**

rS_{n} = ar + ar^{2} + ar^{3} + ...... + ar^{n} ---- (2)

**Step 3: Perform Subtraction**

Subtract the original sum equation (1) from the one obtained by multiplying by r (equation 2):

rS_{n} = ar + ar^{2} + ar^{3} + ...... + ar^{n}

- S_{n} = - (a + ar + ar^{2}+ ..... + ar^{(n - 1)}

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(r - 1)S_{n} = ar^{n} - a

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Or (r - 1)S_{n} = a(r^{n} - 1)

**Step 4: Solve for S _{n}**

Solve the equation for S_{n} by isolating it on one side:

S_{n} = \\frac{a \left (r^n - 1 \right )}{r - 1})

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