Geometric Progressions
Exercise 8.2 Solutions
Q1
1. Find 20th and nth terms of G.P. $\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \dots$
Identify
$a = \frac{5}{2}$
Ratio r
$r = \frac{5/4}{5/2} = \frac{1}{2}$
$a_{20}$
$ar^{19} = \frac{5}{2}(\frac{1}{2})^{19} = \frac{5}{2^{20}}$
$a_n$
$ar^{n-1} = \frac{5}{2}(\frac{1}{2})^{n-1} = \frac{5}{2^1 \cdot 2^{n-1}} = \frac{5}{2^n}$
$a_{20} = \frac{5}{2^{20}}, \quad a_n = \frac{5}{2^n}$
Q2
2. Find 12th term of G.P. whose 8th term is 192 and common ratio is 2.
Given
$a_8 = 192, \quad r = 2$
Formula
$a_8 = ar^7 \Rightarrow a(2^7) = 192$
Find a
$a(128) = 192 \Rightarrow a = \frac{192}{128} = \frac{3}{2}$
$a_{12}$
$ar^{11} = \frac{3}{2}(2)^{11} = 3 \cdot 2^{10} = 3(1024)$
Answer: 3072
Q3
3. 5th, 8th, 11th terms are p, q, s. Show $q^2 = ps$.
Terms
$p = ar^4, \quad q = ar^7, \quad s = ar^{10}$
LHS ($q^2$)
$(ar^7)^2 = a^2 r^{14}$
RHS ($ps$)
$(ar^4)(ar^{10}) = a^2 r^{4+10} = a^2 r^{14}$
LHS = RHS. Proved.
Q4
4. 4th term is square of 2nd term. First term is -3. Find 7th term.
Given
$a_4 = (a_2)^2, \quad a = -3$
Eq
$ar^3 = (ar)^2 \Rightarrow ar^3 = a^2r^2$
Solve r
$r = a = -3$ (dividing by $ar^2$)
$a_7$
$ar^6 = (-3)(-3)^6 = (-3)^7$
Answer: -2187
Q5
5. Which term of sequence: (a) $2, 2\sqrt{2}, 4, \dots$ is 128?
[Image of geometric progression graph]
Setup
$a=2, r=\sqrt{2}$. Let $a_n = 128$.
Eq
$2(\sqrt{2})^{n-1} = 128 \Rightarrow (\sqrt{2})^{n-1} = 64$
Powers
$2^{\frac{n-1}{2}} = 2^6 \Rightarrow \frac{n-1}{2} = 6$
Solve
$n-1 = 12 \Rightarrow n = 13$
13th term
Q6
6. For what x are $-\frac{2}{7}, x, -\frac{7}{2}$ in G.P.?
Condition
$\frac{b}{a} = \frac{c}{b} \Rightarrow b^2 = ac$ (Geometric Mean)
Eq
$x^2 = (-\frac{2}{7})(-\frac{7}{2}) = 1$
Solve
$x = \pm 1$
x = ±1
Q7
7. Sum to 20 terms: $0.15, 0.015, 0.0015, \dots$
Setup
$a=0.15, \quad r=0.1 = \frac{1}{10}$ (since $r < 1$)
Formula
$S_{20} = \frac{a(1-r^{20})}{1-r} = \frac{0.15(1 – 0.1^{20})}{1 – 0.1}$
Simplify
$\frac{0.15}{0.9}(1 – 0.1^{20}) = \frac{15}{90}(1 – 0.1^{20})$
$\frac{1}{6}(1 – (0.1)^{20})$
Q9
9. Sum to n terms: $1, -a, a^2, -a^3, \dots$ ($a \neq -1$)
Identify
$a_1 = 1, \quad r = -a$
Formula
$S_n = \frac{1(1 – (-a)^n)}{1 – (-a)}$
$S_n = \frac{1 – (-a)^n}{1 + a}$
Q11
11. Evaluate $\sum_{k=1}^{11} (2 + 3^k)$
Split
$\sum_{k=1}^{11} 2 + \sum_{k=1}^{11} 3^k$
Part 1
$2 \times 11 = 22$
Part 2
GP: $3, 3^2, \dots, 3^{11}$. ($a=3, r=3, n=11$).
$S = \frac{3(3^{11}-1)}{3-1} = \frac{3}{2}(3^{11}-1)$
$S = \frac{3(3^{11}-1)}{3-1} = \frac{3}{2}(3^{11}-1)$
$22 + \frac{3}{2}(3^{11} – 1)$
Q12
12. Sum of first 3 terms is $\frac{39}{10}$, product is 1. Find terms.
Let
Terms be $\frac{a}{r}, a, ar$.
Product
$\frac{a}{r} \cdot a \cdot ar = 1 \Rightarrow a^3 = 1 \Rightarrow a = 1$.
Sum
$\frac{1}{r} + 1 + r = \frac{39}{10} \Rightarrow \frac{1+r+r^2}{r} = \frac{39}{10}$
Solve
$10r^2 – 29r + 10 = 0 \Rightarrow (2r-5)(5r-2)=0$.
r
$r = \frac{5}{2}$ or $\frac{2}{5}$.
Terms: $\frac{2}{5}, 1, \frac{5}{2}$ or $\frac{5}{2}, 1, \frac{2}{5}$
Q13
13. How many terms of GP $3, 3^2, 3^3 \dots$ give sum 120?
Setup
$a=3, r=3, S_n=120$.
Formula
$120 = \frac{3(3^n – 1)}{3-1} = \frac{3}{2}(3^n – 1)$
Solve
$80 = 3^n – 1 \Rightarrow 3^n = 81$
Power
$3^n = 3^4 \Rightarrow n = 4$
4 terms
Q14
14. Sum of first 3 is 16, sum of next 3 is 128. Find $S_n$.
Eq 1
$S_3 = \frac{a(r^3-1)}{r-1} = 16$
Eq 2
Sum next 3 $= S_6 – S_3 = 128 \Rightarrow S_6 = 144$.
Divide
$\frac{S_6}{S_3} = \frac{r^6-1}{r^3-1} = r^3+1 = \frac{144}{16} = 9$
Ratio r
$r^3 = 8 \Rightarrow r = 2$.
Find a
$\frac{a(8-1)}{1} = 16 \Rightarrow 7a = 16 \Rightarrow a = 16/7$.
$S_n = \frac{16}{7}(2^n – 1)$
Q15
15. Given $a=729, a_7=64$. Find $S_7$.
Find r
$ar^6 = 64 \Rightarrow 729r^6 = 64 \Rightarrow r^6 = \frac{64}{729} = (\frac{2}{3})^6 \Rightarrow r = \frac{2}{3}$
Formula
$S_7 = \frac{a(1-r^7)}{1-r} = \frac{729(1 – \frac{64}{729} \cdot \frac{2}{3})}{1/3}$ (Oops, use $r^7$ logic)
Calc
$S_7 = \frac{729(1 – (2/3)^7)}{1/3} = 2187(1 – \frac{128}{2187})$
Simplify
$2187 – 128 = 2059$
S_7 = 2059
Geometric Progressions
Exercise 8.2 Solutions (Q16 – Q32)
Q16
16. Find a G.P. where sum of first two terms is -4 and 5th term is 4 times the 3rd term.
Condition 1
$a + ar = -4 \Rightarrow a(1+r) = -4 \dots (1)$
Condition 2
$a_5 = 4 a_3 \Rightarrow ar^4 = 4 ar^2$
Solve r
$r^2 = 4 \Rightarrow r = \pm 2$
Case r=2
From (1): $a(1+2) = -4 \Rightarrow 3a = -4 \Rightarrow a = -4/3$.
GP: $-4/3, -8/3, -16/3, \dots$
GP: $-4/3, -8/3, -16/3, \dots$
Case r=-2
From (1): $a(1-2) = -4 \Rightarrow -a = -4 \Rightarrow a = 4$.
GP: $4, -8, 16, -32, \dots$
GP: $4, -8, 16, -32, \dots$
Two possible G.P.s found.
Q17
17. If 4th, 10th, 16th terms are x, y, z. Prove x, y, z in G.P.
Terms
$x = ar^3, \quad y = ar^9, \quad z = ar^{15}$
Check GP
To be in G.P., $y^2$ must equal $xz$.
LHS ($y^2$)
$(ar^9)^2 = a^2 r^{18}$
RHS ($xz$)
$(ar^3)(ar^{15}) = a^2 r^{3+15} = a^2 r^{18}$
LHS = RHS. Hence x, y, z are in G.P.
Q18
18. Sum to n terms: 8, 88, 888, …
Step 1
$S_n = 8[1 + 11 + 111 + \dots]$
Step 2
Multiply/Divide by 9: $\frac{8}{9}[9 + 99 + 999 + \dots]$
Step 3
Rewrite as powers of 10: $\frac{8}{9}[(10-1) + (100-1) + (1000-1) + \dots]$
Step 4
$\frac{8}{9}[(10 + 10^2 + \dots + 10^n) – (1 + 1 + \dots + 1)]$
Sum Formula
$\frac{8}{9}[\frac{10(10^n – 1)}{10-1} – n]$
$\frac{80}{81}(10^n – 1) – \frac{8n}{9}$
Q19
19. Sum of products of corresponding terms: (2, 4, 8, 16, 32) and (128, 32, 8, 2, 1/2).
Products
$2 \times 128 = 256$
$4 \times 32 = 128$
$8 \times 8 = 64$
$16 \times 2 = 32$
$32 \times 0.5 = 16$
$4 \times 32 = 128$
$8 \times 8 = 64$
$16 \times 2 = 32$
$32 \times 0.5 = 16$
Sequence
$256, 128, 64, 32, 16$. This is a G.P.
Parameters
$a = 256, \quad r = 1/2, \quad n = 5$.
Sum
$S_5 = \frac{256(1 – (1/2)^5)}{1 – 1/2} = \frac{256(1 – 1/32)}{1/2}$
Calc
$512 \times \frac{31}{32} = 16 \times 31 = 496$.
Sum = 496
Q20
20. Products of terms of sequences $a, ar, ar^2 \dots$ and $A, AR, AR^2 \dots$. Show G.P.
Terms
$aA, (ar)(AR), (ar^2)(AR^2), \dots$
Pattern
$aA, aA(rR), aA(rR)^2, \dots$
Conclusion
Each term is previous term multiplied by $(rR)$.
It is a G.P. with common ratio rR.
Q21
21. Find 4 numbers in GP. 3rd is greater than 1st by 9. 2nd is greater than 4th by 18.
Eq 1
$ar^2 = a + 9 \Rightarrow a(r^2 – 1) = 9$
Eq 2
$ar = ar^3 + 18 \Rightarrow ar(1 – r^2) = 18$
Divide
$\frac{ar(1-r^2)}{a(r^2-1)} = \frac{18}{9} \Rightarrow -r = 2 \Rightarrow r = -2$.
Find a
$a(4-1) = 9 \Rightarrow 3a = 9 \Rightarrow a = 3$.
Numbers: 3, -6, 12, -24
Q22
22. If p-th, q-th, r-th terms are a, b, c. Prove $a^{q-r} b^{r-p} c^{p-q} = 1$.
Terms
$a = AR^{p-1}, \quad b = AR^{q-1}, \quad c = AR^{r-1}$
Substitute
$(AR^{p-1})^{q-r} (AR^{q-1})^{r-p} (AR^{r-1})^{p-q}$
Powers of A
$A^{(q-r) + (r-p) + (p-q)} = A^0 = 1$
Powers of R
Sum of exponents becomes 0 (Cyclic terms cancel). $R^0 = 1$.
Result is $1 \times 1 = 1$. Proved.
Q23
23. 1st term a, nth term b, Product P. Prove $P^2 = (ab)^n$.
Product P
$a \cdot ar \cdot ar^2 \dots ar^{n-1} = a^n r^{1+2+\dots+(n-1)}$
Simplify P
$a^n r^{\frac{n(n-1)}{2}}$
Square P
$P^2 = a^{2n} r^{n(n-1)} = (a^2 r^{n-1})^n$
RHS
$(ab)^n = (a \cdot ar^{n-1})^n = (a^2 r^{n-1})^n$
LHS = RHS. Proved.
Q24
24. Ratio of sum of first n terms to sum of terms from (n+1) to 2n.
Sum 1
$S_n = \frac{a(1-r^n)}{1-r}$
Sum 2
Terms from $n+1$ start with $ar^n$. No. of terms = n.
Sum = $\frac{ar^n(1-r^n)}{1-r}$
Sum = $\frac{ar^n(1-r^n)}{1-r}$
Ratio
$\frac{S_n}{\text{Sum 2}} = \frac{a(1-r^n)/(1-r)}{ar^n(1-r^n)/(1-r)} = \frac{1}{r^n}$
Ratio is $1/r^n$.
Q25
25. If a, b, c, d in GP, show $(a^2+b^2+c^2)(b^2+c^2+d^2) = (ab+bc+cd)^2$.
Terms
$b=ar, c=ar^2, d=ar^3$
LHS Factor 1
$a^2+a^2r^2+a^2r^4 = a^2(1+r^2+r^4)$
LHS Factor 2
$a^2r^2+a^2r^4+a^2r^6 = a^2r^2(1+r^2+r^4)$
LHS Total
$a^4 r^2 (1+r^2+r^4)^2$
RHS Inner
$a(ar) + ar(ar^2) + ar^2(ar^3) = a^2r + a^2r^3 + a^2r^5 = a^2r(1+r^2+r^4)$
RHS Squared
$(a^2r)^2 (1+r^2+r^4)^2 = a^4r^2(1+r^2+r^4)^2$
LHS = RHS. Proved.
Q26
26. Insert two numbers between 3 and 81 so that resulting sequence is GP.
Setup
$3, G_1, G_2, 81$. Total terms $n=4$.
Find r
$a_4 = ar^3 \Rightarrow 81 = 3r^3 \Rightarrow r^3 = 27 \Rightarrow r = 3$.
Terms
$G_1 = 3(3) = 9$. $G_2 = 9(3) = 27$.
Numbers: 9, 27
Q27
27. Find value of n so $\frac{a^{n+1}+b^{n+1}}{a^n+b^n}$ is geometric mean of a and b.
GM
$\sqrt{ab} = (ab)^{1/2}$
Eq
$\frac{a^{n+1}+b^{n+1}}{a^n+b^n} = a^{1/2}b^{1/2}$
Cross Mult
$a^{n+1}+b^{n+1} = a^{n+1/2}b^{1/2} + a^{1/2}b^{n+1/2}$
Group
$a^{n+1} – a^{n+1/2}b^{1/2} = a^{1/2}b^{n+1/2} – b^{n+1}$
Factor
$a^{n+1/2}(a^{1/2}-b^{1/2}) = b^{n+1/2}(a^{1/2}-b^{1/2})$
Solve
$(\frac{a}{b})^{n+1/2} = 1 = (\frac{a}{b})^0 \Rightarrow n = -1/2$
n = -1/2
Q28
28. Sum of two numbers is 6 times their GM. Show ratio is $(3+2\sqrt{2}):(3-2\sqrt{2})$.
Given
$a+b = 6\sqrt{ab}$
Method
$\frac{a+b}{2\sqrt{ab}} = \frac{3}{1}$. Apply Componendo & Dividendo.
Apply C&D
$\frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}} = \frac{3+1}{3-1} \Rightarrow \frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2} = \frac{4}{2} = 2$
Root
$\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}} = \frac{\sqrt{2}}{1}$
C&D Again
$\frac{2\sqrt{a}}{2\sqrt{b}} = \frac{\sqrt{2}+1}{\sqrt{2}-1} \Rightarrow \frac{a}{b} = (\frac{\sqrt{2}+1}{\sqrt{2}-1})^2 = \frac{2+1+2\sqrt{2}}{2+1-2\sqrt{2}}$
Ratio: $(3+2\sqrt{2}) : (3-2\sqrt{2})$
Q29
29. If A and G are AM and GM, prove numbers are $A \pm \sqrt{(A+G)(A-G)}$.
Given
$a+b = 2A, \quad ab = G^2$
Quadratic
Equation with roots a, b: $x^2 – (2A)x + G^2 = 0$
Solve
$x = \frac{2A \pm \sqrt{4A^2 – 4G^2}}{2} = A \pm \sqrt{A^2 – G^2}$
Simplify
$A \pm \sqrt{(A+G)(A-G)}$
Proved.
Q30
30. Bacteria doubles every hour. Initial 30. Count at 2nd, 4th, nth hour?
Sequence
30, 60, 120, 240… (GP with $a=30, r=2$)
2nd Hour
Term after 2 hours is $a_3 = 30(2^2) = 120$.
4th Hour
Term after 4 hours is $a_5 = 30(2^4) = 480$.
nth Hour
Term after n hours is $a_{n+1} = 30(2^n)$.
120, 480, $30(2^n)$
Q31
31. Rs 500 at 10% annual interest compounded annually for 10 years.
Formula
$A = P(1 + \frac{R}{100})^n$
Values
$P=500, R=10, n=10$.
Calc
$A = 500(1 + 0.1)^{10} = 500(1.1)^{10}$
Rs $500(1.1)^{10}$
Q32
32. AM=8, GM=5. Find quadratic equation.
Sum
$\frac{a+b}{2} = 8 \Rightarrow a+b = 16$.
Product
$\sqrt{ab} = 5 \Rightarrow ab = 25$.
Equation
$x^2 – (\text{Sum})x + (\text{Product}) = 0$
$x^2 – 16x + 25 = 0$