# Geometric Progression

A sequence is said to be in Geometric Progression, if the ratio between any two adjacent numbers in the sequence is constant (non zero). This constant is said to be common ratio (c.r.)

e.g. 1, 2, 4, 8 ..........â€¦â€¦ c.r. = 2

1, 1/2, 1/4, 1/8, â€¦..â€¦â€¦â€¦ c.r. =1/2

1/4, â€“1/2, 1, â€“2, 4, â€¦â€¦â€¦ c.r. = â€“2

In general a, ar, ar^{2}, ar^{3}, â€¦â€¦, a r^{nâ€“1}

By having a close look we can say the n-th term of GP will be given by t_{n} = ar^{nâ€“1}

Sum of a geometric progression

S_{n} = a + ar + ar^{2} +â€¦+ ar^{nâ€“1} â€¦ (6)

Multiplying both sides by r, we get,

rS_{n} = ar + ar^{2} +â€¦+ ar^{nâ€“1} + ar^{n} â€¦ (7)

Subtracting (7) from (6), we have

S_{n} â€“ rS_{n} = a â€“ ar^{n}

or, S_{n} = a(1â€“r^{r})/(1â€“r) â€¦ (8)

Thus

S_{n} = a(1â€“r^{n})/(1â€“r)

What happens when n tends to infinity.

If we take any value of |r| greater than 1 then value of r^{n} when n â†’âˆž will tend to infinite. Hence value of S_{âˆž} will also tend to infinite. If we take of |r| less than 1 then r^{n }will tend to 0 when n â†’ âˆž .

Thus S_{âˆž} = a/1â€“r for |r| < 1 â€¦ (9)

When any real number is added, subtracted, multiplied or divided to each term of a geometric series?

1. Multiplication/Division by a constant number to each term of a G.P. also results a G.P.

Suppose a_{1}, a_{2}, a_{3}, â€¦â€¦, a_{n} are in G.P.

then ka_{1}, ka_{2}, ka_{3}, â€¦â€¦, ka_{n} and

a_{1}/k, a_{2}/k, ... ... ... a_{n}/k will also be in G.P.

Where k ? R and k â‰ 0.

2. Multiplication/Division of two G.P.â€™s also results a G.P.

Suppose a_{1}, a_{2}, a_{3}, â€¦â€¦, a_{n}

and b_{1}, b_{2}, b_{3}, â€¦â€¦, b_{n} are two G.P.

then a_{1}b_{1}, a_{2}b_{2}, a_{3}b_{3}, â€¦â€¦, a_{n }b_{n}

then a_{1}/b_{2}, a_{2}/b_{2}, ... ... ..., a_{n}/b_{n} will also be in G.P.

3. Reversing the order of a G.P.â€™s also results a G.P.

Suppose a_{1}, a_{2}, a_{3}, â€¦â€¦, a_{n} are in G.P.

then a_{n}, a_{nâ€“1}, a_{nâ€“1}, â€¦â€¦, a_{3}, a_{2}, a_{1} will also be in G.P.

4. Taking the inverse of a G.P. also results a G.P.

Suppose a_{1}, a_{2}, a_{3}, â€¦â€¦, a_{n} are in G.P.

then 1/a_{1}, 1/a_{2}, 1/a â€¦â€¦, 1/a_{n} will also be in G.P.

To solve problems numbers should be taken intelligently eg.

Three number in G.P. âˆ´ Î±/ÃŸ, Î±, Î±ÃŸ c.r = ÃŸ

Four number in G.P. âˆ´ Î±/ÃŸ^{3}, Î± ÃŸ, Î±ÃŸ, Î±ÃŸ^{3} c.r. = ÃŸ^{2}

Five numbers in G.P. âˆ´ Î±/ÃŸ^{2}, Î±/ÃŸ, Î± . Î±ÃŸ . Î±ÃŸ^{2} c.r. = ÃŸ

â€¢ If each term of a G.P. is multiplied (or divided) by a fixed non-zero constant, then the resulting sequence is also a G.P. with same ratio as that of the given G.P.

â€¢ If each term of a G.P. (with common ratio r) is raised to the power k, then the resulting sequence is also a G.P. with common ratio r^{k}.

â€¢ If a_{1}, a_{2}, a_{3}, â€¦â€¦, b_{1}, b_{2}, b_{3}, â€¦â€¦ are two G.P.â€™s with common ratios r and râ€™ respectively then the sequence a_{1}b_{1}, a_{2}b_{2}, a_{3}b_{3}, â€¦â€¦ is also a G.P. with common ratio rrâ€™.

â€¢ If we have to take three terms in a G.P., it is convenient to take them as a/r, a, ar. In general, we take a/r^{k}, a/r^{kâ€“1}, â€¦, a, a^{r}, â€¦, ar^{k} in case we have to take (2k + 1) terms in a G.P.

â€¢ If we have to take four terms in a G.P., it is convenient to take them as a/r^{3}, a/r, ar, ar^{3}. In general, we take a/r^{2kâ€“1}, a/r^{2kâ€“3}, ... ... ... a/r, ar, â€¦â€¦, ar^{2kâ€“1}, in case we have to take 2k terms in a G.P.

â€¢ If a_{1}, a_{2}, â€¦â€¦, a_{n} are in G.P., then a_{1}a_{n} = a_{2}a_{nâ€“1} = a_{3}a_{nâ€“2} = â€¦â€¦

â€¢ If a_{1}, a_{2}, a_{3}, â€¦â€¦ is a G.P. (each a_{1} > 0), then loga_{1}, loga_{2}, loga_{3} â€¦â€¦ is an A.P. The converse is also true.

# Geometric Mean between two numbers

If three numbers are in G.P. then middle one is said to be geometric mean (GM) of two others.

Let, G be the geometric mean between two number a and b

So, a G b are in G.P.

G/a = b/G.

or, G^{2} = ab

âˆ´ G =âˆšab

Similarly we can find two geometric means between two given numbers a and b.

Let a, G_{1}, G_{2}, b are in G.P.

t_{n} = a r^{nâ€“1}

or b/a = r^{3}

r =(b/a)1/3

G_{1} = ar^{2} = a (b/a)^{1/3} = a^{1/3} b^{2/3}

# Geometric Mean(s)

â€¢ If three terms are in G.P., then the middle term is called the geometric mean (G.M.) between the two. So if a, b, c are in G.P., then b = âˆšac is the geometric mean of a and c.

â€¢ If a_{1}, a_{2}, â€¦â€¦, a_{n} are non-zero positive numbers, then their G.M.(G) is given by G = (a_{1}a_{2}a_{3}, â€¦â€¦, a_{n})^{1/n}. If G_{1}, G_{2}, â€¦â€¦ G_{n} are n geometric means between and a and b then a, G_{1}, G_{2}, â€¦â€¦, G_{n} b will be a G.P. Here b = ar^{n+1}.

â‡’ r = ^{n+1}âˆšb/a â‡’ G_{1} = a^{n+1}âˆšb/a, G_{2} = a(^{n+1}âˆšb/a)^{2},â€¦, G_{n} = a(^{n+1}âˆšb/a)^{n}.

The 7^{th} term of a G.P. is 8 times the 4^{th} term. Find the G.P. when its 5^{th} term is 48.

Given that t_{7} = 8t_{4} â‡’ ar^{6} = 8ar^{3}

â‡’ r^{3} = 8 = 23 â‡’ r = 2.

Also t_{5} = 48 â‡’ ar^{4} = 48 or 16a = 48 â‡’ a = 3.

Hence the required G.P. is 3, 6, 12, 24 â€¦â€¦

Does there exists a G.P. containing 27, 8 and 12 as three of its terms? If it exists, how many such progressions are possible?

Let 8 be the m^{th}, 12 the n^{th} and 27 be the t^{th} terms of a G.P. whose first term is A and common ratio is R.

Then 8 = AR^{mâ€“1}, 12 = AR^{nâ€“1}, 27 = AR^{tâ€“1}

â‡’ 8/12 = R^{mâ€“n} = 2/3, 12/27 = R^{nâ€“t} = (2/3)^{2}, 8/27 = R^{mâ€“t} =(2/3)^{3}

â‡’ 2m â€“ 2n = n â€“ t and 3m â€“ 3n = m â€“ t

â‡’ 2m + t = 3n and 2m + t = 3n

â‡’ 2m+t/3 = n.

â€‹There are infinite sets of values of m, n, t which satisfy this relation. For example, take m = 1, then 2+t/3 = n = k â‡’ n = k, t = 3k â€“ 2. By giving different values to k we get integral values of n and t. Hence there are infinite numbers of G.P.â€™s whose terms may be 27, 8, 12 (not consecutive).

In a four term series if first three are in G.P. and last three are in A.P. with common different 6 and last terms is equal to the first term then find all four terms in series.

This is very tricky question. If you read question carefully then it is clear that we have to start with A.P. because common difference is given.

Let the numbers be a + 6, aâ€“6, a, a+6 now first three are in G.P. is (aâ€“6)^{2} = a(a+6) or, a^{2} â€“ 12a + 36 = a^{2} therefore numbers are 8, â€“4, 2, 8.