A geometric progression is a sequence where the succeeding term is ‘r’ times the preceding term. If a paper is folded four or five times, calculating the height of the stack of the paper after being folded four to five times is an example of geometric progression. In this article, we come across geometric progression solved problems along with definition and properties of common ratio.
Geometric Progression Definition
A sequence in which the first term being non zero and the ratio of each term to its previous term remains a constant. The ratio is represented by the letter “r” and is called the common ratio.
The properties of a geometric progression are:
- If a constant non zero quantity is multiplied or divided to each term, the resulting sequence is also a geometric progression.
- The reciprocal of all the terms in geometric progression also form a geometric progression.
- If all the terms in a geometric progression are raised to the same power, then the new sequence obtained is also in geometric progression.
Finite and infinite geometric progression
The geometric progression with a finite number of terms is a finite geometric progression whereas an infinite number of terms is an infinite geometric progression.
Important Results
The formulae related to a geometric progression are given below:
Representation of GP | a, ar, ar^{²}, ar^{³}, ……… ar^{n-1}, ar^{n} | a = first term r = common ratio n = the last term in GP and,S_{n} = sum to n terms of GP |
The common ratio of GP | ||
The n^{th} term of GP | a_{n} = ar^{n-1} | |
Sum to n terms of a GP | S_{n} = a (1−r^{n}) / 1−r |
To Determine Whether The Given Sequence Is A GP Or Not
If the common ratio between the terms is a constant, then it is said to be a geometric sequence.
For instance : Check whether the sequence 8, 16, 32, 64, 128 . . . . . is a GP.
Solution:
Given
Since the common ratio is a constant, it is a GP.
Properties Of Common Ratio [r]
- The value of the common ratio determines whether the sequence is increasing or decreasing.
- Positive: The common ratio r being positive leads to the first term of GP being positive and thus GP to be positive.
- Negative: The common ratio r being negative leads to the first term of GP being positive and thus GP to be negative.
- r > 1 indicates that the GP has exponential growth towards positive infinity.
- r < -1 indicates that the GP has exponential growth towards infinity in the alternating sign.
- If the value of r lies between -1 and 1, then GP has exponential decay towards 0.
Related articles
Solve Problems on Geometric Progression for IIT JEE
Example 1: If the 4^{th}, 7^{th} and 10^{th} terms of a G.P. be a, b, c respectively, then what is the relation between a, b, c?
Solution:
Let first term of G.P. = A and common ratio = r
We know that n^{th }term of G.P. = Ar^{n−1 }
Now t_{4 }= a = Ar^{3}, t_{7 }= b = Ar^{6} and t_{10}= c = Ar^{9}
Relation b^{2 }= ac is true because
B^{2 }= (Ar^{6})^{2 }= A^{2}r^{12} and ac = (Ar^{3}) * (Ar^{9}) = A^{2}r^{12}
As we know, if xy + 2y^{2} + yz = xy + xz + y^{2} + yz are in A.P., then
2n^{2} + 5n − 2n^{2 }+ 4n − 2 − 5n + 5 = 4n + 3 terms of a G.P. are always in G.P.,
Therefore, a, b, c will be in G.P. i.e. 2, 5, 8, 11, 14 = 40.
Example 2: If a, b, c are p^{th}, q^{th} and r^{th} terms of a G.P., then what is (c / b)^{p }(b / a)^{r }(a / c)^{q} equal to?
Solution:
a = AR^{p−1}, b = AR^{q−1}, c = AR^{r−1}
(c / b)^{p }* (b / a)^{r }* (a / c)^{q }= [(AR^{r−1} / AR^{q−1}) ]^{p} * [(AR^{q−1} / AR^{p−1})]^{r } * [(AR^{p−1}/ AR^{r−1})]^{q}
= R^{(r−q)p+(q−p)r+(p−r)q}
= R^{0}
= 1
Example 3: The 6^{th} term of a G.P. is 32 and its 8^{th} term is 128, then find the common ratio of the G.P.
Solution:
T_{6 }= 32 and T_{8 }= 128
⇒ ar_{5 }= 32 ….. (i) and
ar_{7 }= 128 …..(ii)
Dividing (ii) by (i), we have
r^{2}=4
r = 2
Example 4: Find the sum of the series 6 + 66 + 666 + ……….upto n terms.
Solution:
Given series 6 + 66 + 666 + ……….upto n terms
= [6 / 9] (9 + 99 + 999 + ….. upto n terms)
= [2 / 3] (10 + 10^{2} + 10^{3 }+ ………. + upto n terms −n)
= [2 / 3] [10 (10^{n}−1) / 10−1] −n)
= [1 / 27] [20(10^{n }− 1) − 18n]
= 2 (10^{n+1}− 9n − 10) / 27
Example 5: The sum of a few terms of any ratio series is 728, if the common ratio is 3 and the last term is 486, then what will be the first term of series?
Solution:
n^{th} term of series = ar^{n−1 }= a(3)^{n−1 }= 486 —– (i)
Sum of n terms of series is S_{n}= a (3^{n}−1) / [3−1] = 728 (∵r >1)
From (i), a (3^{n} / 3) = 486 or [a * 3^{n}] = 3 × 486 = 1458
From (ii), [a * 3^{n}] − a = 728 * 2 or [a * 3^{n}] − a
1458 − a = 1456
or a = 2
Example 6: If three geometric means be inserted between 2 and 32, then find the third geometric mean.
Solution:
The GP will be 2, g_{1} , g_{2}, g_{3}, 32 where a = 2, a_{r }= g_{1}, ar_{2}= g_{2}, ar_{3 }= g_{3} and ar_{4 }= 32
Now 2 × r^{4 }= 32
⇒ r^{4 }= 16 = (2)^{4}
⇒ r = 2
Then third geometric mean = ar_{3 }= 2 × 2^{3} = 16
Example 7: An increasing GP is formed by positive numbers. The new numbers form an AP if the middle term of the geometric progression is doubled. Find the common ratio of the GP.
Solution:
Let
If the middle term of the GP is doubled, then the new numbers are in AP.
Since the series is an increasing GP,
Example 8: If
Solution:
Now x, y and z are in GP.
Now GM > HM
Now, for three numbers
Example 9:Find 9th and nth term of a G.P 4,16,64 …
Solution:
Here a = 4
r = 16/4 = 4
a_{9} = ar^{n-1} = 4×4^{9-1} = 4×4^{8} = 49
Also a_{n} = ar^{n-1} = 4(4)^{n-1} = 4^{n}.
Example 10: Find sum of first 5 terms of the G.P 2, 8, 32,…
Solution:
Here a = 2
r = 8/2= 4
n = 5
Sn = a(r^{n}-1)/(r-1)
= 2(4^{5} -1)/(4-1)
= 2(4^{5} -1)/3
= 682
Hence the required sum is 682.