## Geometric Sequence Formulas:

A sequence of non-zero numbers is said to be a geometric sequence if the ratio of any term to its succeeding term remains constant. This constant is called the common ratio of the geometric sequence denoted by ‘r’ i.e.

r = a_{n} / a_{n-1} |

Geometric Sequence is also called *Geometric*** Progression**.

*General Form of Geometric Sequence:*

If a be the first term and r be the common ratio of the geometric sequence, then its general form is a, ar, ar^{2}. ar^{3}, ……..

*n ^{th}– term of Geometric Sequence:*

If a be the first term and r be the common ratio of the geometric sequence, then its n^{th} term is given by-

a_{n }= ar^{n-1} |

*Selection of terms in**Geometric**Sequence:*

Number of Terms | Terms to be Selected | Common Ratio |
---|---|---|

3 | a/r, a, ar | r |

4 | a/r^{3}, a/r, ar, ar^{3} | r^{2} |

5 | a/r^{2}, a/r, a, ar, ar^{2} | r |

*Sum to n-Terms of a**Geometric**Sequence:*

Let a be the first term and r, be the common ratio of a Geometric Sequence. If S_{n} denotes the sum of the first n terms, then

S_{n} = a + ar + ar^{2} + … + ar^{n-1}∴ r . S _{n} = ar + ar^{2} + ar^{3} + … + ar^{n-1} + ar^{n}Subtracting, S _{n} (1 – r) = a – ar^{n}⇒ S _{n} (1 – r) = a (1 – r^{n})⇒ ………..(i)S_{n} = a (1 – r^{n}) / (1 – r)This formula is suitable for finding out the sum of n terms in the Geometric Sequence when |r| < 1. But if |r| > 1, then we write, ⇒ ………..(ii)S_{n} = a (r^{n} – 1) / (r – 1)If r = 1, then the above formulae fail to find out the sum. But in this case, S _{n} = a + a + a + … + to n terms = na |

*Sum to Infinite Terms of a**Geometric**Sequence:*

S_{∞} = a/(1 – r) [r < 1] |

Example-Find the eighth term and the general term of the geometric sequence 9, 18, 36, … The given geometric sequence is,Solution-9, 18, 36, … Here, a = 9, r = 18/9 = 2 Now, a _{n }= ar^{n-1}⇒ a _{n }= 9 (2)^{n-1}and a _{8} = 9 (2)^{7}⇒ a _{8} = 9 (128)⇒ a _{8} = 1152 |

Example-The third term of a G.P. is 12. Find the product of its first five terms. Let ‘a’ and ‘r’ be the first term and common ratio of G.P. such that- Solution-a _{3} = 12⇒ ar ^{2} = 12 ……….(i)Now, the product of the first five terms of G.P. = (a) (ar) (ar ^{2}) (ar^{3}) ar^{4})= a ^{5}r^{1+2+3+4+5}= a ^{5}r^{10}= (ar ^{2})^{5}= 12 ^{5} |

Example-If l, m, and n be the p^{th}, q^{th }and r^{th} terms of a G.P. respectively, show that l^{q-r} m^{r-p} n^{p-q} = 1. Let ‘a’ and ‘rSolution-_{1}‘ be the first term and common ratio of a G.P. respectively.Now, a _{p} = l⇒ ar _{1}^{p-1} = lAlso a _{q} = m⇒ ar _{1}^{q-1} = m and ar = n ⇒ ar _{1}^{r-1} = nNow, l ^{q-r} m^{r-p} n^{p-q}= [ar _{1}^{p-1}]^{q-r }[ar_{1}^{q-1}]^{r-p }[ar_{1}^{r-1}]^{p-q}= a ^{q-r} . r_{1}^{(p-1) (q-r)} . a^{r-p} . r_{1}^{(q-1) (r-p)} . a^{p-q} . r_{1}^{(r-1) (p-q)}= a ^{q-r+r-p+p-q} . r_{1}^{pq-pr-q+r+qr-qp-r+p+rp-rq-p+q}= a ^{0} . r_{1}^{0}= 1 |

Example-The third, seventh, and eleventh terms of a geometric sequence are x, y and z respectively. Show that y^{2} = xz. Let ‘a’ and ‘r’ be the first term and common ratio of a geometric mean respectively.Solution-Now, a _{3} = x⇒ ar ^{2} = x ……….(i)Also a _{7} = y⇒ ar ^{6} = y ……….(ii)and a _{11} = z⇒ ar ^{10} = z ……….(ii)Now, xz = (ar ^{2}) (ar^{10})⇒ xz = a ^{2}r^{12}⇒ xz = (ar ^{6})^{2}⇒ xz = y ^{2} |

Example- If a^{x} = b^{y}= c^{z} and x, y, z are in geometric sequence, prove that log_{b}a = log_{c}b. aSolution-^{x} = b^{y} = c^{z} and x, y, z are in geometric sequence.y ^{2} = xz⇒ x = y ^{2}/z ……….(i)Now, a ^{x}= b^{y} = c^{z}Apply log log a ^{x} = log b^{y} = log c^{z}⇒ x log a = y log b = z log c = k ⇒ log a = k/x, log b = k/y and log c = k/z Now, log _{b}a = log a/log b = (k/x)/(k/y) = y/x = y/(y ^{2}/z) [using (i)]= z/y ……….(A) Now, log _{c}b = log b/log c = (k/y)/(k/z) = z/y ……….(B) From (A) and (B):- log _{b}a = log_{c}b |

Example-If a, b, c are in a geometric sequence, prove that 1/log_{a }m, 1/log_{b }m, and 1/log_{c} m are in an arithmetic sequence. a, b, c are in a geometric sequenceSolution-⇒ b ^{2} = acApply log ⇒ log b ^{2} = log ac⇒ 2log _{m }b = log_{m} a + log_{m} c⇒ log _{m} a, log_{m }b, log_{m} c are in an arithmetic sequence.⇒ 1/log _{a} m, 1/log_{b }m, 1/log_{c} m are in an arithmetic sequence. |

Example- If l, m, n are in A.P., show that the l^{th}, m^{th} and n^{th} terms of any G.P. are also in G.P. l, m, n are in A.P.Solution-⇒ 2m = l + n ⇒ m + m = l + n ⇒ m – l = n – m ……….(i) Let ‘a’ and ‘r’ be the first term and common ratio of a geometric sequence. Now, a _{l} = ar^{l-1}a _{m} = ar^{m-1}and a _{n} = ar^{n-1}Now, a _{m}/a_{l} = ar^{m-1}/ar^{l-1}⇒ a _{m}/a_{l} = r^{m-l} ……….(ii)and a _{n}/a_{m} = ar^{n-1}/ar^{m-1}⇒ a _{n}/a_{m} = r^{n-m} ……….(ii)From (i), (ii) and (iii):- a _{m}/a_{l} = a_{n}/a_{m}⇒ (am)2 = (al) (an) ⇒ a _{l}, a_{m}, a_{n} are in a geometric sequence. |

Example-Find the fourth term from the end of the G.P. 6, 12, 24, …….. 24576. The given G.P. isSolution-6, 12, 24, …….. 24576 or 24576, 12288, …….., 24, 12, 6 Here a = 24576, r = 1/2 Now, a _{4} = ar^{3}= 24576 (1/2) ^{3}= 24576/8 = 3072 |

Example-The sum of three numbers in G.P. is 65 and their product is 3375. Find the numbers. Let three numbers in G.P. be Solution-a/r, a, a _{r}Now, (a/r) (a) (ar) = 3375 ⇒ a ^{3} = 3375⇒ a ^{3} = 15^{3}⇒ a = 15 Also, a/r + a + ar = 65 ⇒ a (1/r + 1 + r) = 65 ⇒ 15 (1 + r + r ^{2})/r = 65⇒ 3 + 3r + 3r ^{2} = 13r⇒ 3r ^{2} – 10r + 3 = 0⇒ r = 3, 1/3 ∴ three numbers in G.P. are 5, 15, 45, or 45, 15, 5. |

Example- If x, 2x + 2, 3x + 3 are the first three terms of a G.P., find its fourth term. x, 2x + 2, 3x + 3 are first three terms of G.P.Solution-⇒ (2x + 2) ^{2} = x (3x + 3)⇒ 4(x + 1) ^{2} = 3x (x + 1)⇒ 4(x + 1) = 3x ⇒ 4x + 4 = 3x ⇒ x = -4 ∴ three terms in G.P. are -4, 2(-4) + 2, 3(-4) + 3 = -4, -6, -9 Here r = -6/-4 = 3/2 ∴ a _{n }= ar^{3}⇒ a _{n }= (-4) (3/2)^{3}⇒ a _{n }= -4 x 27/8⇒ a _{n }= -27/2 |

Example- How many terms of the geometric sequence 6, 3, 3/2 ……… are needed to give the sum 3069/256? The given geometric sequence is,Solution-6, 3, 3/2 ……… Here a = 6, r = 3/6 = 1/2 Let S _{n} = 3069/256⇒ a(r ^{n} – 1)/(r – 1) = 3069/256⇒ 6[(1/2) ^{n} – 1]/[(1/2) – 1] = 3069/256⇒ 6[(1/2) ^{n} – 1]/(-1/2) = 3069/256⇒ -12 [(1/2) ^{n} – 1] = 3069/256⇒ 12 (1 – 1/2 ^{n}) = 3069/256⇒ 1 – 1/2 ^{n} = 3069/256 x 12⇒ 1 – 1/2 ^{n} = 1023/1024⇒ 1 – 1023/1024 = 1/2 ^{n}⇒ 1/1024 = 1/2 ^{n}⇒ 1/2 ^{10} = 1/2^{n}⇒ n = 10 |

Example-Find the sum to n terms 6 + 66 + 666 + ……… Let SSolution-_{n} = 6 + 66 + 666 + ……… to n terms⇒ S _{n} = 6 [1 + 11 + 111 + ……… to n terms]⇒ S _{n} = 6/9 [9 + 99 + 999 + ……… to n terms]⇒ S _{n} = 6/9 [(10 – 1] + (100 – 1) + (1000 – 1) + ……… to n terms] ⇒ S _{n} = 6/9 [(10 + 100 + 1000 + ……… to n terms) – n]⇒ S _{n} = 6/9 [(sum of n terms of G.P. with a = 10, r = 10) – n]⇒ S _{n} = 6/9 [{10 (10^{n} – 1) / (10 – 1)} – n]⇒ S _{n} = 6/9 [10 (10^{n} – 1)/9 – n]⇒ S _{n} = (6/9) [10 (10^{n} – 1) – 9n]/9⇒ S _{n} = (6/81) (10^{n+1} – 10 – 9n)⇒ S _{n} = (2/27) (10^{n+1} – 9n – 10) |

Example-If the fourth and the seventh terms of a series in G.P. be 24 and 192 respectively, find the sum of its first twelve terms. Let ‘a’ and ‘r’ be the first term and common ratio of a G.P. respectively such that Solution-a _{4} = 24⇒ ar ^{3} = 24 ……….(i)and a _{7} = 192⇒ ar ^{6} = 192 ……….(ii)Dividing (i) by (ii):- ar ^{6}/ar^{3} = 192/24⇒ r ^{3} = 8⇒ r = 2 Substitute r = 2 in equation (i):- a(2) ^{3} = 24⇒ 8a = 24 ⇒ a = 3 ∴ S _{n} = a (r^{12} – 1) / (r – 1)⇒ S _{n} = 3(2^{12} – 1)/(2 – 1)⇒ S _{n} = 3(4096 – 1)/1⇒ S _{n} = 3(4095)⇒ S _{n} = 12285 |