Geometric Progression (G.P.) — Real-Life Applications in Finance, Insurance & Science
Geometric Progression (G.P.) is a key mathematical concept where each term is obtained by multiplying the previous term by a fixed number called the common ratio. It is not just theory — it plays a major role in finance, insurance, investments, and real-world growth models.
In daily life, we see geometric progressions in compound interest, bank savings plans, population growth, car depreciation, and radioactive decay. The same principle also applies in calculating insurance premiums, loan repayment schedules, and financial growth forecasts.
In business and economics, understanding G.P. helps you model exponential changes — whether predicting stock market returns, evaluating credit score impact, or optimizing retirement savings plans. In science, G.P. appears in radioactive decay, light intensity, and nuclear chain reactions.
Studying G.P. improves your ability to:
- Predict long-term growth using compound interest and investment calculators.
- Compare insurance coverage, loan plans, and savings over time.
- Understand how repeated multiplication drives exponential changes in finance and nature.
Geometric Progression — Key Formulas
Let the terms be \(t_1,t_2,\dots,t_n\) with first term \(a=t_1\) and common ratio \(r\). Below are the important formulas you will use.
- General (nth) term: \[ t_n = a r^{\,n-1}. \]
- Common ratio (from consecutive terms): \[ r = \frac{t_n}{t_{n-1}}, \quad \text{when }t_{n-1}\neq0. \]
- Sum of first \(n\) terms (finite sum), \(r\ne1\): \[ S_n = a\left(\frac{1-r^n}{1-r}\right). \]
- Sum of first \(n\) terms (finite sum), \(r\geq 1\): \[ S_n = a\left(\frac{r^n-1}{r-1}\right). \]
- If \(r=1\): \[ S_n = na. \]
- Sum from the \(m\)-th to the \(n\)-th term: \[ \sum_{k=m}^n ar^{k-1} = a r^{m-1}\cdot\frac{1-r^{\,n-m+1}}{1-r}. \]
- Infinite sum (convergent) — \(|r|<1 b=""> \[ S_{\infty} = \frac{a}{1-r}, \quad |r|<1 .="" li="">
- Product of first \(n\) terms: \[ P_n = a^n r^{\frac{n(n-1)}{2}}. \]
- Geometric mean of first \(n\) terms: \[ G_n = a\,r^{\frac{n-1}{2}}. \]
Tip: Remember both equivalent forms of \(S_n\): \(\dfrac{1-r^n}{1-r}\) and \(\dfrac{r^n-1}{r-1}\) — choose based on \(r\)’s value to avoid sign mistakes.
Q1). Find the common ratio of the geometric progression \(2, 6, 18, 54, \dots\).
Solution:
Formula for common ratio: \(r = \frac{t_2}{t_1}\)
⇒ \(r = \frac{6}{2} = 3\)
Q2). Find the common ratio of the geometric progression \(81, 27, 9, 3, \dots\).
Solution:
Formula for common ratio: \(r = \frac{t_2}{t_1}\)
⇒ \(r = \frac{27}{81} = \frac{1}{3}\)
Q3). Find the 7th term of the geometric progression \(2, 6, 18, 54, \dots\)
Solution:
In a geometric progression (G.P.), the nth term is given by \(t_n = a r^{n-1}\).
Here, \(a = 2\) and \(r = \frac{6}{2} = 3.\)
Therefore, the 7th term is:
\(t_7 = 2(3)^{7-1} = 2 \times 3^6 = 2 \times 729 = 1458.\)
✅ Answer: The 7th term is \(1458.\)
Q4). The 5th and 8th terms of a G.P. are 48 and 384 respectively. Find the geometric progression.
Solution:
Let \(a\) be the first term and \(r\) be the common ratio.
Then, \(t_5 = a r^4 = 48\) and \(t_8 = a r^7 = 384.\)
Dividing the equations:
\(\frac{t_8}{t_5} = \frac{a r^7}{a r^4} = r^3 = \frac{384}{48} = 8 \Rightarrow r = 2.\)
Substitute \(r = 2\) in \(a r^4 = 48\):
\(a(2)^4 = 48 \Rightarrow 16a = 48 \Rightarrow a = 3.\)
Hence, the required G.P. is:
\(3, 6, 12, 24, 48, 96, 192, 384, \dots\)
✅ Answer: \(a = 3, \ r = 2.\)
Q5). How many terms of the G.P. \(3, 6, 12, 24, \dots\) are required to give a sum of 1533?
Solution:
In a G.P., the sum of n terms is \(S_n = a \frac{r^n - 1}{r - 1}\).
Here, \(a = 3,\ r = 2,\ S_n = 1533.\)
Substitute values:
\(1533 = 3 \times \frac{2^n - 1}{2 - 1}\)
\(\Rightarrow 2^n - 1 = 511\)
\(\Rightarrow 2^n = 512.\)
Since \(2^9 = 512,\ n = 9.\)
✅ Answer: The sum of the first 9 terms is 1533.
Q6). A bacteria culture doubles every hour. If there are 200 bacteria now, find the population after 5 hours.
Solution:
This is a G.P. where each term doubles the previous one.
Hence, \(a = 200,\ r = 2.\)
The number after 5 hours is the 6th term (including the initial count):
\(t_6 = a r^{6-1} = 200(2)^5 = 200 \times 32 \)
\(= 6400.\)
✅ Answer: After 5 hours, the population will be 6400 bacteria.
Q7). Find the common ratio and write the next four terms of each G.P.
(i) \(0.5,\;0.05,\;0.005,\dots\)
Solution:
Next four terms = \(0.0005,\;0.00005,\;0.000005,\;0.0000005.\)
✅ Answer: \(r=0.1;\) next four terms: \(5\times10^{-4},\;5\times10^{-5},\;5\times10^{-6},\;5\times10^{-7}.\)
(ii) \(2,\;-6,\;18,\;-54,\dots\)
Solution:
Next four terms = \(162,\;-486,\;1458,\;-4374.\)
✅ Answer: \(r=-3;\) next four terms: \(162,\;-486,\;1458,\;-4374.\)
Q8 (ii). The 5th and 8th terms of a G.P. are 48 and 384 respectively. Find the geometric progression.
Solution:
Dividing: \(r^3 = \frac{384}{48} = 8 \Rightarrow r = 2.\)
Substitute: \(a(2^4)=48 \Rightarrow a=3.\)
✅ Answer: \(a=3,\ r=2,\) G.P. = \(3,6,12,24,48,96,192,384,\dots\)
Q9). Derive the formula for the sum of the first \(n\) terms of a G.P. and state the result when \(r=1.\)
Solution:
\(rS_n = ar + ar^2 + \dots + ar^n\)
Subtract: \(S_n - rS_n = a - ar^n\)
\(S_n(1 - r) = a(1 - r^n)\)
\(\boxed{S_n = a\frac{1 - r^n}{1 - r} = a\frac{r^n - 1}{r - 1}}\)
If \(r=1\), then \(S_n = na.\)
✅ Answer: \(S_n = a\frac{1 - r^n}{1 - r}\) for \(r \ne 1;\) if \(r=1,\ S_n=na.\)
Q10). Find the 7th term of the geometric progression \(3,\,6,\,12,\,24,\dots\)
Solution:
\(t_7 = 3(2^6) = 3 \times 64 = 192.\)
✅ Answer: The 7th term is \(192.\)
Q11). Find the 9th term of the geometric progression \(3,\,6,\,12,\,24,\dots\)
Solution:
\(t_9 = a r^{9-1} = 3(2^8) = 3 \times 256 = 768.\)
✅ Answer: The 9th term is \(768.\)
Q12). If the p-th term of a G.P. is \(P\) and the q-th term is \(Q\), show that the n-th term is \[ a_n \;=\; \Bigg(\dfrac{P^{\,n-q}}{\,Q^{\,n-p}}\Bigg)^{\!1/(p-q)}. \]
Full solution and explanation:
1. Set up the general G.P. notation.
Let the first term be \(A\) and the common ratio be \(r\). Then the \(k\)-th term of the G.P.
is given by the standard formula
\[
a_k \;=\; A\,r^{\,k-1}.
\]
In particular:
\[
P = a_p = A\,r^{\,p-1},\qquad
Q = a_q = A\,r^{\,q-1}.
\]
These two equations express the given data in terms of the unknowns \(A\) and \(r\).
2. Eliminate the first term \(A\) by dividing the two equations.
Divide \(Q\) by \(P\):
\[
\frac{Q}{P} \;=\; \frac{A\,r^{\,q-1}}{A\,r^{\,p-1}}
\;=\; r^{\,q-p}.
\]
This step removes \(A\) and yields an equation involving only \(r\) and known quantities \(P,Q,p,q\).
Hence the common ratio is
\[
r \;=\; \Big(\frac{Q}{P}\Big)^{\!1/(q-p)}.
\]
(This requires \(q\neq p\); if \(q=p\) the statement is degenerate because two different term-numbers
cannot give different term values in a nontrivial G.P.)
3. Express the n-th term in terms of P and r (eliminating A).
From \(P = A r^{p-1}\) we can solve for \(A\):
\[
A \;=\; P\,r^{-(p-1)}.
\]
Substitute this into the general \(n\)-th term formula \(a_n = A r^{n-1}\):
\[
a_n \;=\; \big(P\,r^{-(p-1)}\big)\,r^{\,n-1}
\;=\; P\,r^{\,n-p}.
\]
So the \(n\)-th term can be written compactly as \(a_n = P\,r^{\,n-p}\).
(Equivalently, using \(Q\) you would get \(a_n = Q\,r^{\,n-q}\).)
4. Substitute the expression for \(r\) in terms of \(P,Q\).
Replace \(r\) by \(\big(\tfrac{Q}{P}\big)^{1/(q-p)}\) in \(a_n = P\,r^{\,n-p}\):
\[
a_n
\;=\; P \left(\Big(\frac{Q}{P}\Big)^{\!1/(q-p)}\right)^{\,n-p}
\;=\; P^{\,1 - \frac{n-p}{q-p}}\; Q^{\,\frac{n-p}{q-p}}.
\]
This algebraic manipulation moves the fractional exponent inside and separates powers of \(P\) and \(Q\).
5. Rewrite the result in the requested symmetric form.
The expression above is algebraically equivalent to several compact forms. One convenient symmetric
form is obtained by rearranging exponents:
\[
a_n \;=\; P^{\frac{q-n}{q-p}}\; Q^{\frac{n-p}{q-p}}
\;=\; \Bigg(\dfrac{P^{\,n-q}}{\,Q^{\,n-p}}\Bigg)^{\!1/(p-q)}.
\]
\[ \boxed{\;a_n \;=\; \Bigg(\dfrac{P^{\,n-q}}{\,Q^{\,n-p}}\Bigg)^{\!1/(p-q)}\;} \]
Example 13). Find the sum to n terms of the series :
\(7 + 77 + 777 + \ldots\)
Solution.
\(= 7(1 + 11 + 111 + \ldots \text{ to } n \text{ terms})\)
\(= \frac{7}{9}(9 + 99 + 999 + \ldots \text{ to } n \text{ terms})\)
\(= \frac{7}{9}[(10 - 1) + (100 - 1) + (1000 - 1) + \ldots \text{ to } n \text{ terms}]\)
\(= \frac{7}{9}[(10 + 100 + 1000 + \ldots \text{ to } n \text{ terms})\)
\(- (1 + 1 + 1 + \ldots \text{ to } n \text{ terms})]\)
\(= \frac{7}{9}\left[\frac{10(10^n - 1)}{10 - 1} - n\right]\)
\(= \frac{7}{81}(10^{n+1} - 10 - 9n)\)
Sum of first 50 terms:
\(= \frac{7}{81}(10^{51} - 460)\)
Example 14. Determine the number of terms \(n\) in a geometric progression if \(a = 3,\ a_n = 96,\ \text{and}\ S_n = 189.\)
The n-th term of a G.P. is \(a_n = a r^{n-1}\).
Substituting given values: \(96 = 3 r^{n-1}\)
\(\Rightarrow r^{n-1} = 32.\)
Hence \(r = 2,\) because \(32 = 2^5.\)
Now, use the sum formula \(S_n = a \frac{r^n - 1}{r - 1}\):
\[
189 = 3\frac{2^n - 1}{2 - 1} \Rightarrow 63 = 2^n - 1.
\]
Therefore \(2^n = 64 \Rightarrow n = 6.\)
✅ Answer: Number of terms \(n = 6.\)
Example 15. Find the sum of the series \(11 + 103 + 1005 + \dots\) up to \(n\) terms.
Solution:
We can separate each term into two parts: a geometric and an arithmetic component.\(11 = (10 + 1),\; 103 = (10^2 + 3),\;\)
\(1005 = (10^3 + 5),\; \dots\)
So, \[ S_n = (10 + 10^2 + 10^3 + \dots + 10^n)\] \[+ (1 + 3 + 5 + \dots + (2n - 1)). \] The first part is a G.P. with first term \(10\), ratio \(10\), and \(n\) terms: \[ \sum 10^k = 10\frac{10^n - 1}{9}. \] The second part is an A.P. with first term \(1\), common difference \(2\), and \(n\) terms: \[ \sum (2k - 1) = \frac{n}{2}[2 + (n - 1)\times 2] = n^2. \] Combining both parts: \[ S_n = \frac{10(10^n - 1)}{9} + n^2. \] ✅ Answer: \(S_n = \dfrac{10(10^n - 1)}{9} + n^2.\)
Example 16. Find the least value of n for which the sum \(1 + 3 + 3^2 + \dots + 3^{\,n-1}\) is greater than \(7000.\)
Solution:
Example 17. Sum to n terms :
\(4 + 44 + 444 + \ldots\)
Solution.
\(= 4(1 + 11 + 111 + \ldots \text{ to } n \text{ terms})\)
\(= \frac{4}{9}(9 + 99 + 999 + \ldots \text{ to } n \text{ terms})\)
\(= \frac{4}{9}[(10 - 1) + (100 - 1) + (1000 - 1) + \ldots \text{ to } n \text{ terms}]\)
\(= \frac{4}{9}[(10 + 100 + 1000 + \ldots \text{ to } n \text{ terms})\)
\(- (1 + 1 + 1 + \ldots \text{ to } n \text{ terms})]\)
\(= \frac{4}{9}\left[\frac{10(10^n - 1)}{10 - 1} - n\right]\)
\(= \frac{4}{81}(10^{n+1} - 10 - 9n)\)
Example 18. Use geometric series to express \(0.555\ldots\) as a rational number.
Solution. Here,
Thus \(0.555\ldots\) is the sum of an infinite G.P. having
first term \(a = 0.5\) and \(r = 0.1.\)
Since \(|r| < 1,\) we can find the sum
\[ S = \frac{a}{1-r} = \frac{0.5}{1-0.1} = \frac{0.5}{0.9} = \frac{5}{9}. \] Thus \(0.555\ldots = \frac{5}{9},\) which is a rational number.
Example 19. Evaluate \(0.2345\overline{45}\).
Solution. Here,
\(+ 0.00000045 + \ldots\)
\( = 0.23 + \frac{45}{1000}\left(1 + \frac{1}{100} + \frac{1}{100^2} + \frac{1}{100^3} + \ldots\right)\)
\( = 0.23 + \frac{45}{1000}\left[\frac{1}{1 - \frac{1}{100}}\right]\)
\( = 0.23 + \frac{45}{1000} \times \frac{100}{99}\)
\( = \frac{23}{100} + \frac{45}{990}\)
\( = \frac{2277 + 450}{9900} = \frac{2727}{9900} = \frac{303}{1100}.\)
Hence, \(0.2345\overline{45} = \frac{303}{1100}.\)
19). (Standard GP form) Consider the sequence \(1,\; -a,\; a^2,\; -a^3,\; \dots\) (first term \(1\), common ratio \(-a\)). Find the sum of the first \(n\) terms (assume \(a\neq -1\)).
This is a geometric progression with first term \(A=1\) and common ratio \(r=-a\).
The sum of first \(n\) terms of a GP is
\[
S_n = A\frac{1-r^{n}}{1-r}.
\]
Substituting \(A=1\) and \(r=-a\) gives
\[
\boxed{\,S_n = \frac{1-(-a)^{n}}{1-(-a)} = \frac{1-(-a)^{n}}{1+a}\, ,\quad (a\neq -1).}
\]
(If you prefer, split by parity of \(n\): when \(n\) is even \((-a)^n=a^n\), when \(n\) is odd \((-a)^n=-a^n\).)
20). How many terms of the G.P. \(3,\; \tfrac32,\; \tfrac34,\; \dots\) are needed to give the sum \(\dfrac{3069}{512}\)?
Here \(a=3\) and common ratio \(r=\tfrac12\). The sum of first \(n\) terms is \[ S_n = a\frac{1-r^n}{1-r} = 3\cdot\frac{1-(\tfrac12)^n}{1-\tfrac12}\] \[= 6\Big(1-\Big(\tfrac12\Big)^n\Big). \] Set \(S_n=\dfrac{3069}{512}\): \[ 6\Big(1-\Big(\tfrac12\Big)^n\Big) = \frac{3069}{512} \quad\] \[\Rightarrow\quad 1-\Big(\tfrac12\Big)^n = \frac{3069}{3072}. \] So \[ \Big(\tfrac12\Big)^n = 1 - \frac{3069}{3072} = \frac{3}{3072} = \frac{1}{1024}. \] Hence \(2^{-n} = 2^{-10}\), so \(\boxed{\,n=10\,}\).
21). If \(\{a_n\}\) is a G.P. with \(a=4\) and \(r=5\), find \(a_6\) and \(S_6\).
The nth term is \(a_n = a r^{\,n-1}\). Thus \[ a_6 = 4\cdot 5^{5} = 4\cdot 3125 = 12500. \] The sum of first 6 terms (since \(r\ne1\)) is \[ S_6 = a_1\frac{r^{6}-1}{r-1} = 4\cdot\frac{5^{6}-1}{4} = 5^{6}-1 = 15625-1\[ \[= 15624. \] \(\boxed{\,a_6=12500,\quad S_6=15624\,.}\)
22). Determine the number \(n\) of terms of the G.P. \(3,6,12,\dots\) so that \(S_n=381\).
Here \(a=3,\ r=2\). Sum formula:
\[
S_n = a\frac{r^{n}-1}{r-1} = 3\frac{2^{n}-1}{1} = 3(2^{n}-1).
\]
Set \(3(2^{n}-1)=381\Rightarrow 2^{n}-1=127\)
\(\Rightarrow 2^{n}=128\)
\(\Rightarrow \boxed{\,n=7\,}.\)
23). (Short answer) The sum of some terms of a G.P. is \(315\). The first term is \(5\) and the common ratio is \(2\). Find the last term and the number of terms.
Let the number of terms be \(n\). Sum formula:
\[
S_n = a\frac{r^{n}-1}{r-1} = 5\frac{2^{n}-1}{1} = 5(2^{n}-1).
\]
Given \(5(2^{n}-1)=315\Rightarrow 2^{n}-1=63\)
\(\Rightarrow 2^{n}=64\Rightarrow n=6.\)
Last term \(t_n = a r^{\,n-1} = 5\cdot 2^{5} = 5\cdot 32 = \boxed{160}.\)
Exercise 24). Find the sum to \(n\) terms of the series:
\(9 + 99 + 999 + \cdots\)
Solution:
Each term is formed by repeating 9’s — we can express every term as a combination of powers of 10.
Step 1 : Write the general term \(t_k\):
\[ t_k = 9 + 90 + 900 + \dots + 9\!\times\!10^{\,k-1}. \] Here the first term is \(9\) and the common ratio is \(10\). Hence the term is a finite geometric sum: \[ t_k = 9\,\frac{10^{\,k}-1}{10-1} = 9\,\frac{10^{\,k}-1}{9} = 10^{\,k}-1. \] Step 2 : Write the total sum of the first \(n\) terms: \[ S_n = t_1 + t_2 + t_3 + \dots + t_n \] \[ = (10^1 - 1) + (10^2 - 1) + (10^3 - 1) + \dots \] \[+ (10^n - 1). \] Step 3 : Separate into two sums: \[ S_n = (10 + 10^2 + 10^3 + \dots + 10^n) \] \[- (1 + 1 + 1 + \dots + 1). \] The second sum has \(n\) ones, so it equals \(n\): \[ S_n = (10 + 10^2 + 10^3 + \dots + 10^n) - n. \] Step 4 : The first part is a geometric progression with first term \(10\), ratio \(10\), and \(n\) terms: \[ \sum_{k=1}^{n} 10^k = 10\,\frac{10^{\,n}-1}{10-1} = \frac{10(10^{\,n}-1)}{9}. \] Step 5 : Substitute this into the expression for \(S_n\): \[ S_n = \frac{10(10^{\,n}-1)}{9} - n = \frac{10^{\,n+1}-10-9n}{9}. \]
\(\displaystyle S_n = \frac{10^{\,n+1}-10-9n}{9}.\)
Q25). Prove that \(9^{1/3} \cdot 9^{1/9} \cdot 9^{1/27} \cdot \ldots \text{ to } \infty = 3.\)
Solution:
The given product is \[ 9^{\frac{1}{3}} \times 9^{\frac{1}{9}} \times 9^{\frac{1}{27}} \times \cdots = 9^{\left(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots\right)}. \] The exponents form a G.P. with first term \(a=\frac{1}{3}\) and ratio \(r=\frac{1}{3}\). Sum to infinity: \[ S = \frac{a}{1-r} = \frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{1}{2}. \] Therefore, \[ 9^{S} = 9^{1/2} = 3. \]Q26). The common ratio of a G.P. is \(-\tfrac{4}{5}\) and the sum to infinity is \(\tfrac{80}{9}\). Find the first term.
Solution:
Formula for sum to infinity (\(|r|<1 -="" 16.="" a="\frac{80}{9}" and="" class="boxed" div="" frac="" hence="" infty="\frac{80}{9}\)" r="-\frac{4}{5}\):" s_="" substitute="" times="">Answer: First term \(a = 16.\)Q27). Find \(S_\infty\) of a G.P. whose first term is \(28\) and the fourth term is \(\tfrac{4}{49}.\)
Solution:
Since \(|r|< 1\)
\(S_\infty = \frac{a}{1-r} =\frac{28}{1-\dfrac{1}{7}}= \dfrac{98}{3}.\)
Q28). Evaluate \(1.\overline{36}\)
Solution:
We can separate the integer part and the repeating decimal part: \[ x = 1 + 0.\overline{36}. \] Now focus on the repeating portion \(0.\overline{36}\): \[ 0.\overline{36} = 0.36 + 0.0036 + 0.000036 + \ldots \] Here the first term \(a = 0.36\) and the common ratio \(r = 0.01.\)
Since \(|r| < 1\), we can use the formula for the sum of an infinite geometric series: \[ S_\infty = \frac{a}{1 - r}. \] Substituting values: \[ S_\infty = \frac{0.36}{1 - 0.01} = \frac{0.36}{0.99} = \frac{36}{99}. \] Simplify: \[ \frac{36}{99} = \frac{4}{11}. \] Hence, \[ x = 1 + \frac{4}{11} = \frac{11 + 4}{11} = \frac{15}{11}. \]
29). Sum to \(n\) terms:
\(0.3 + 0.33 + 0.333 + \ldots\)
Solution:
\( = \frac{1}{3}[(1 - 0.1) + (1 - 0.01) + (1 - 0.001) + \ldots \text{ to } n \text{ terms}] \)
\( = \frac{1}{3}[(1 + 1 + 1 + \ldots \text{ to } n \text{ terms}) \)
\(- (0.1 + 0.01 + 0.001 + \ldots \text{ to } n \text{ terms})] \)
\( = \frac{1}{3}\left[n - \left(\frac{1}{10} + \frac{1}{100} + \frac{1}{1000} + \ldots \text{ to } n \text{ terms}\right)\right] \)
\( = \frac{1}{3}\left[n - \frac{\frac{1}{10}\left(1 - \left(\frac{1}{10}\right)^n\right)}{1 - \frac{1}{10}}\right] \)
\( = \frac{1}{3}\left[n - \frac{1}{9}\left(1 - \frac{1}{10^n}\right)\right] \)
\( = \frac{1}{27}\left[9n - 1 + \frac{1}{10^n}\right] \)
Q1). In a nuclear chain reaction, one neutron strikes an atom and releases 2 more neutrons. Each of these new neutrons again causes fission and releases 2 more neutrons in the next collision. Find the number of neutrons released after the 10th collision.
Explanation & Solution:
Let the first neutron that starts the reaction be the first term, \[ a = 1 \] and each neutron produces two new ones, so the common ratio \[ r = 2. \] The number of neutrons after the 2nd, 3rd, 4th, ... collisions will be: \[ 1,\; 2,\; 4,\; 8,\; 16,\; 32,\; \dots \] This is a geometric sequence with ratio 2. The number of neutrons after the 10th collision is the 10th term: \[ t_{10} = a \times r^{n-1} = 1 \times 2^{10-1} = 2^9 = 512. \] ✅ Answer: There will be 512 neutrons after the 10th collision.
Hence, the rapid multiplication of neutrons in fission follows a geometric growth pattern.
Q2). A person sends a letter to two friends. Each of these friends sends a copy of the same letter to two more friends, and the process continues in this way. Find the total number of letters sent after 10 rounds.
Explanation & Solution:
Here, the number of letters in each round forms a G.P.: \[ a = 1,\quad r = 2. \] So, the letters sent in each round are: \[ 1,\; 2,\; 4,\; 8,\; 16,\; 32,\; \dots \] The total number of letters sent up to the 10th round will be the sum of 10 terms: \[ S_{10} = a \frac{r^{10} - 1}{r - 1} = \frac{2^{10} - 1}{1} = 1023. \] ✅ Answer: A total of 1,023 letters will be sent after 10 rounds.
Word Problem
Q3 — Bouncing ball. A ball is dropped from height 100 m. Each bounce reaches 80% of the previous height. Find the total distance travelled before it comes to rest.
First drop = \(100\) m. Rebound heights form a G.P. with first rebound \(h_1=100\times0.8=80\) m and common ratio \(r=0.8\).
After the first drop the ball goes up \(h_1\) and down \(h_1\), then up \(h_2\) and down \(h_2\), etc. So total distance \[ D = \text{initial drop} + 2\sum_{k=1}^{\infty} h_k. \]
\(h_k = 100\cdot(0.8)^k\) for \(k\ge1\). Sum of rebound heights: \[ \sum_{k=1}^{\infty} h_k = 100\sum_{k=1}^{\infty}(0.8)^k\)\] \[ = 100\cdot\frac{0.8}{1-0.8} = 100\cdot\frac{0.8}{0.2} = 100\cdot4 = 400. \]
\[ D = 100 + 2\times 400 = 100 + 800 = 900\ \text{m}. \]
Q4 — Car depreciation. A new car costs ₹8,00,000 and depreciates by 15% each year. Find its value after 5 years.
Depreciation 15% per year ⇒ value each year multiplies by \(1-0.15=0.85\). So \(r=0.85\).
If \(V_0=800000\), then after \(n\) years \[ V_n = V_0 \cdot r^n = 800000\cdot(0.85)^n. \]
Exact expression: \[ V_5 = 800000\cdot(0.85)^5. \] Numeric approximation: \((0.85)^2=0.7225,\ (0.85)^3\approx0.614125,\)
\(\ (0.85)^4\approx0.522006,\ (0.85)^5\approx0.443705.\)
\[ V_5 \approx 800000\times 0.443705\] \[\approx 3,54,964\ (\text{approx}). \]
Q5 — Compound interest (bank deposit). A man deposits ₹5,000; bank pays 8% compounded yearly. Find the amount after 6 years.
Annual multiplier \(=1+0.08=1.08\).
\[ A_n = P(1+r)^n = 5000\cdot(1.08)^6. \]
Compute \((1.08)^6\) (approx): \(1.08^2=1.1664,\ 1.08^3\approx1.259712,\ 1.08^4\)
\(\approx1.36049,\ 1.08^5\approx1.46933,\ 1.08^6\approx1.58687.\) \[ A_6 \approx 5000\times1.58687 \approx ₹7,934.35. \]
Q6 — Pendulum swings. A pendulum swings 10 cm on the 1st oscillation, 9 cm on the 2nd, 8.1 cm on the 3rd, … (each term 0.9 of previous). Find total distance covered in 15 oscillations.
Given amplitudes \(a_1=10\) cm and ratio \(r=0.9\). One full oscillation goes from centre to one side and back to centre and to other side then back — but commonly "one oscillation" here is counted as a single to-and-fro swing; the distance travelled in one oscillation ≈ \(2\times\) (amplitude for that oscillation). (We follow the textbook convention: distance per oscillation = \(2a_k\).)
Amplitudes: \(a_k = 10(0.9)^{\,k-1}\). Sum of first 15 amplitudes: \[ \sum_{k=1}^{15} a_k = 10\sum_{k=0}^{14}(0.9)^k = 10\cdot\frac{1-(0.9)^{15}}{1-0.9}. \]
\[ D_{15} = 2\sum_{k=1}^{15} a_k = 20\cdot\frac{1-(0.9)^{15}}{0.1}\] \[ = 200\big(1-(0.9)^{15}\big)\ \text{cm}. \]
\((0.9)^{15}\approx 0.2059\) (approx), so \[ D_{15}\approx 200(1-0.2059)=200\times0.7941\approx 158.82\ \text{cm}. \]
Q7 — Radioactive decay (half-life). Half-life = 10 hours. Initial quantity 100 g. Find remaining after 50 hours.
After each 10 hours quantity multiplies by \(1/2\). For \(t\) hours: \[ Q(t) = Q_0\left(\tfrac{1}{2}\right)^{t/10}. \]
\(50/10=5\) half-lives, so \[ Q(50) = 100\left(\tfrac{1}{2}\right)^5 = 100\cdot\frac{1}{32} = \frac{100}{32}\] \[= 3.125\ \text{g}. \]
📊 Geometric Progression in Finance — Real-Life Examples & Explanations
Financial growth and decline are common examples of Geometric Progression (G.P.). Whenever money increases or decreases by a fixed percentage — such as in compound interest, stock investments, insurance premiums, or depreciation — it follows a geometric pattern. Below are five real-world finance examples showing how G.P. helps calculate future or total values.
Example 1: Compound Interest — Bank Savings
Problem: A person deposits ₹10,000 at 10% annual interest compounded yearly. Find the amount after 5 years.
Explanation: In compound interest, each year’s amount becomes the next year’s principal — a perfect geometric sequence.
\( t_5 = a r^{5} = 10000(1.1)^5 = ₹16105.10 \).
✅ Answer: The balance after 5 years = ₹16,105.10.
Each year’s growth is 10%, showing exponential (geometric) increase.
Example 2: Depreciation — Car Value
Problem: A car bought for ₹8,00,000 loses 15% of its value every year. Find its value after 4 years.
Explanation: Depreciation is reverse compound interest — each year the value becomes a fixed fraction of the previous one.
\( t_4 = 800000(0.85)^4 = ₹417600 \).
✅ Answer: The car’s value after 4 years = ₹4,17,600.
Here G.P. shows constant percentage loss instead of gain.
Example 3: Insurance Premium Growth
Problem: An insurance company increases a policy’s annual premium by 5% each year. If the first-year premium is ₹20,000, find it after 6 years.
Explanation: Each year’s premium depends on last year’s premium multiplied by 1.05, forming a G.P.
\( t_6 = 20000(1.05)^5 = ₹25,525.60 \).
✅ Answer: 6th-year premium = ₹25,525.60.
In real insurance models, such growth predicts long-term policy cost.
Example 4: Investment in Stocks
Problem: An investor buys shares worth ₹5,000 that appreciate 12% annually. Find the value after 8 years.
Explanation: Market growth compounds annually, creating geometric increase in portfolio value.
\( t_8 = 5000(1.12)^8 = ₹12,380 \).
✅ Answer: Value after 8 years = ₹12,380.
This shows how consistent returns multiply investment exponentially.
Example 5: Retirement Savings Plan
Problem: A person deposits ₹1,000 yearly in a retirement plan that grows by 8% annually. Find the total amount after 10 years.
Explanation: Each deposit grows geometrically because older deposits earn interest for more years than newer ones.
\( S_{10} = a\frac{r^{10}-1}{r-1} = 1000\frac{(1.08)^{10}-1}{0.08} = ₹14,486 \).
✅ Answer: Total savings after 10 years = ₹14,486.
G.P. helps calculate compound value of regular contributions — vital for retirement planning.
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