Sequences and Series
Miscellaneous Exercise Solutions (Q1 – Q10)
Q1
1. If $f(x+y) = f(x)f(y)$, $f(1)=3$, $\sum_{x=1}^n f(x) = 120$. Find n.
Identify
Since $f(x+y)=f(x)f(y)$, $f(x)$ is exponential $a^x$.
Base
$f(1) = a^1 = 3 \Rightarrow f(x) = 3^x$. This forms a G.P.
Sum
$\sum_{x=1}^n 3^x = 3 + 3^2 + \dots + 3^n = 120$.
Formula
$\frac{3(3^n – 1)}{3 – 1} = 120 \Rightarrow \frac{3}{2}(3^n – 1) = 120$.
Solve
$3^n – 1 = \frac{120 \times 2}{3} = 80 \Rightarrow 3^n = 81 \Rightarrow n = 4$.
n = 4
Q2
2. Sum of G.P. is 315, $a=5, r=2$. Find last term and n.
Sum Eq
$S_n = \frac{a(r^n – 1)}{r – 1} \Rightarrow 315 = \frac{5(2^n – 1)}{2 – 1}$.
Solve n
$315 = 5(2^n – 1) \Rightarrow 63 = 2^n – 1 \Rightarrow 2^n = 64 \Rightarrow n = 6$.
Last Term
$a_6 = ar^5 = 5(2)^5 = 5(32) = 160$.
Last Term = 160, n = 6
Q3
3. First term of G.P. is 1. Sum of 3rd and 5th term is 90. Find common ratio.
Given
$a = 1, \quad a_3 + a_5 = 90$.
Equation
$ar^2 + ar^4 = 90 \Rightarrow r^2 + r^4 = 90$.
Let $x = r^2$
$x + x^2 = 90 \Rightarrow x^2 + x – 90 = 0$.
Factor
$(x+10)(x-9) = 0 \Rightarrow x = 9$ (since $r^2 \neq -10$).
Solve r
$r^2 = 9 \Rightarrow r = \pm 3$.
Common Ratio $r = \pm 3$
Q4
4. Sum of 3 numbers in G.P. is 56. Subtract 1, 7, 21 to get A.P. Find numbers.
GP Terms
$a, ar, ar^2$. Sum: $a(1+r+r^2) = 56 \dots (1)$.
AP Terms
$a-1, \quad ar-7, \quad ar^2-21$.
AP Cond
$2(ar-7) = (a-1) + (ar^2-21) \Rightarrow 2ar – 14 = a + ar^2 – 22$.
Simplify
$a + ar^2 – 2ar = 8 \dots (2)$.
Subtract
Eq(1) – Eq(2): $(a+ar+ar^2) – (a+ar^2-2ar) = 56 – 8$.
$3ar = 48 \Rightarrow ar = 16$.
$3ar = 48 \Rightarrow ar = 16$.
Solve
Sub $a=16/r$ into (1): $\frac{16}{r}(1+r+r^2) = 56 \Rightarrow 2r^2 – 5r + 2 = 0$.
$r = 2$ or $1/2$.
$r = 2$ or $1/2$.
Numbers: (8, 16, 32) or (32, 16, 8)
Q5
5. GP with even terms ($2n$). Sum all = 5 * Sum odd terms. Find r.
Sum All
$S_{all} = \frac{a(r^{2n}-1)}{r-1}$.
Odd Terms
$a, ar^2, ar^4 \dots$ ($n$ terms). First term $a$, ratio $r^2$.
$S_{odd} = \frac{a((r^2)^n – 1)}{r^2 – 1}$.
$S_{odd} = \frac{a((r^2)^n – 1)}{r^2 – 1}$.
Equation
$\frac{a(r^{2n}-1)}{r-1} = 5 \cdot \frac{a(r^{2n}-1)}{r^2-1}$.
Solve
Cancel common terms: $\frac{1}{r-1} = \frac{5}{(r-1)(r+1)} \Rightarrow r+1 = 5 \Rightarrow r = 4$.
Common Ratio r = 4
Q6
6. If $\frac{a+bx}{a-bx} = \frac{b+cx}{b-cx} = \frac{c+dx}{c-dx}$, show a, b, c, d in G.P.
Comp & Div
Apply Componendo & Dividendo to each ratio:
$\frac{(a+bx)+(a-bx)}{(a+bx)-(a-bx)} = \frac{2a}{2bx} = \frac{a}{bx}$.
$\frac{(a+bx)+(a-bx)}{(a+bx)-(a-bx)} = \frac{2a}{2bx} = \frac{a}{bx}$.
Result
Thus, $\frac{a}{bx} = \frac{b}{cx} = \frac{c}{dx}$.
Simplify
Cancel $x$: $\frac{a}{b} = \frac{b}{c} = \frac{c}{d}$.
Invert
$\frac{b}{a} = \frac{c}{b} = \frac{d}{c}$. This is definition of G.P.
Hence a, b, c, d are in G.P.
Q7
7. Prove $P^2 R^n = S^n$ for a G.P.
Sum S
$S = \frac{a(r^n-1)}{r-1}$.
Product P
$P = a^n r^{\frac{n(n-1)}{2}}$. So $P^2 = a^{2n} r^{n(n-1)}$.
Reciprocal R
Sum of $\frac{1}{a}, \frac{1}{ar}, \dots$ (GP with ratio $1/r$).
$R = \frac{\frac{1}{a}(1 – \frac{1}{r^n})}{1 – \frac{1}{r}} = \frac{r^n-1}{a r^{n-1} (r-1)}$.
$R = \frac{\frac{1}{a}(1 – \frac{1}{r^n})}{1 – \frac{1}{r}} = \frac{r^n-1}{a r^{n-1} (r-1)}$.
Compare
Divide $S/R$: $\frac{S}{R} = a^2 r^{n-1}$.
Thus $S^n = R^n (a^2 r^{n-1})^n = R^n P^2$.
Thus $S^n = R^n (a^2 r^{n-1})^n = R^n P^2$.
Hence Proved: $S^n = P^2 R^n$
Q8
8. If a, b, c, d in GP, prove $(a^n+b^n), (b^n+c^n), (c^n+d^n)$ in GP.
Terms
$b=ar, c=ar^2, d=ar^3$.
Check GP
Does $(b^n+c^n)^2 = (a^n+b^n)(c^n+d^n)$?
Simplify
Use substitution $b=ar \dots$. Easier way: $b^n+c^n = r^n(a^n+b^n)$ and $c^n+d^n = r^n(b^n+c^n)$.
Ratio
$\frac{b^n+c^n}{a^n+b^n} = r^n$ and $\frac{c^n+d^n}{b^n+c^n} = r^n$.
Ratios equal. Hence in G.P.
Q9
9. Roots $a,b$ of $x^2-3x+p=0$. Roots $c,d$ of $x^2-12x+q=0$. a,b,c,d in GP. Prove $(q+p):(q-p)=17:15$.
Sums
$a+b=3, \quad c+d=12$.
Ratio
$\frac{c+d}{a+b} = \frac{ar^2(1+r)}{a(1+r)} = r^2 = \frac{12}{3} = 4 \Rightarrow r=2$ (assuming inc).
Products
$p=ab = a(ar) = a^2r$. $q=cd = (ar^2)(ar^3) = a^2r^5$.
Calc
$p = a^2(2) = 2a^2$. $q = a^2(32) = 32a^2$.
Ratio
$\frac{q+p}{q-p} = \frac{32a^2+2a^2}{32a^2-2a^2} = \frac{34a^2}{30a^2} = \frac{17}{15}$.
Ratio 17:15. Proved.
Q10
10. Ratio of AM:GM is $m:n$. Show $a:b = m + \sqrt{m^2-n^2} : m – \sqrt{m^2-n^2}$.
Given
$\frac{a+b}{2\sqrt{ab}} = \frac{m}{n}$.
Comp & Div
$\frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}} = \frac{m+n}{m-n}$. (Apply to $m/n \times 2$ or directly?)
Correct C&D
$\frac{a+b}{2\sqrt{ab}} = \frac{m}{n} \Rightarrow \frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}} = \frac{m+n}{m-n}$.
Roots
$\frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2} = \frac{m+n}{m-n} \Rightarrow \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}} = \sqrt{\frac{m+n}{m-n}}$.
C&D Again
$\frac{2\sqrt{a}}{2\sqrt{b}} = \frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}}$.
Square
$\frac{a}{b} = \frac{m+n + m-n + 2\sqrt{m^2-n^2}}{m+n + m-n – 2\sqrt{m^2-n^2}} = \frac{2m + 2\sqrt{m^2-n^2}}{2m – 2\sqrt{m^2-n^2}}$.
Ratio: $m + \sqrt{m^2-n^2} : m – \sqrt{m^2-n^2}$
Sequences and Series
Miscellaneous Exercise Solutions (Q11 – Q18)
Q11
11. Find the sum to n terms:
(i) $5 + 55 + 555 + \dots$
(ii) $.6 + .66 + .666 + \dots$
(i) $5 + 55 + 555 + \dots$
(ii) $.6 + .66 + .666 + \dots$
(i) $5 + 55 + 555 + \dots$
Step 1
Take 5 common: $5 [1 + 11 + 111 + \dots]$
Step 2
Multiply/Divide by 9: $\frac{5}{9} [9 + 99 + 999 + \dots]$
Step 3
Rewrite terms as powers of 10: $\frac{5}{9} [(10-1) + (100-1) + (1000-1) + \dots]$
Group
$\frac{5}{9} [(10+10^2+\dots+10^n) – (1+1+\dots+1)]$
Result
$\frac{5}{9} [\frac{10(10^n-1)}{10-1} – n] = \frac{50}{81}(10^n-1) – \frac{5n}{9}$
(ii) $0.6 + 0.66 + 0.666 + \dots$
Step 1
$6 [0.1 + 0.11 + 0.111 + \dots]$
Step 2
$\frac{6}{9} [0.9 + 0.99 + 0.999 + \dots] = \frac{2}{3} [(1-0.1) + (1-0.01) + \dots]$
Group
$\frac{2}{3} [(1+1+\dots+n) – (\frac{1}{10} + \frac{1}{100} + \dots)]$
Result
$\frac{2}{3} [n – \frac{1/10(1 – (1/10)^n)}{1 – 1/10}] = \frac{2}{3}[n – \frac{1}{9}(1 – \frac{1}{10^n})]$
(i) $\frac{50}{81}(10^n-1) – \frac{5n}{9}$
(ii) $\frac{2}{3}n – \frac{2}{27}(1 – 10^{-n})$
(ii) $\frac{2}{3}n – \frac{2}{27}(1 – 10^{-n})$
Q12
12. Find the 20th term of the series $2 \times 4 + 4 \times 6 + 6 \times 8 + \dots$
Pattern
1st term: $2 \times 4$. 2nd: $4 \times 6$. 3rd: $6 \times 8$.
nth Term
First factors: $2, 4, 6 \dots (2n)$. Second factors: $4, 6, 8 \dots (2n+2)$.
Formula
$a_n = (2n)(2n+2) = 4n(n+1) = 4n^2 + 4n$.
20th Term
$a_{20} = 4(20)^2 + 4(20) = 4(400) + 80 = 1600 + 80$.
Answer: 1680
Q13
13. Farmer buys tractor for Rs 12000. Pays 6000 cash. Balance paid in annual instalments of Rs 500 + 12% interest on unpaid amount. Total cost?
Balance
Unpaid amount = $12000 – 6000 = 6000$.
Instalments
$\frac{6000}{500} = 12$ instalments.
Interest
1st yr: $12\%$ of 6000 = 720.
2nd yr: $12\%$ of 5500 = 660.
…
12th yr: $12\%$ of 500 = 60.
2nd yr: $12\%$ of 5500 = 660.
…
12th yr: $12\%$ of 500 = 60.
Sum Int
This is an A.P.: $720, 660, \dots, 60$. ($n=12$).
Sum = $\frac{12}{2}(720 + 60) = 6(780) = 4680$.
Sum = $\frac{12}{2}(720 + 60) = 6(780) = 4680$.
Total Cost
Paid Cash (6000) + Instalments Principal (6000) + Total Interest (4680).
Total Cost: Rs 16,680
Q14
14. Scooter cost Rs 22000. Pays 4000 cash. Balance in instalments of Rs 1000 + 10% interest. Total cost?
Balance
$22000 – 4000 = 18000$.
Terms
$\frac{18000}{1000} = 18$ instalments.
Interest
AP Sequence: $10\%$ of 18000, 17000… 1000.
$1800, 1700, \dots, 100$. ($n=18$).
$1800, 1700, \dots, 100$. ($n=18$).
Sum Int
Sum = $\frac{18}{2}(1800 + 100) = 9(1900) = 17100$.
Total
Price (22000) + Interest (17100).
Total Cost: Rs 39,100
Q15
15. Chain letter: Person sends to 4, they send to 4 each. Cost 50 paise/letter. Find cost of 8th set.
Sequence
1st set: 4 letters. 2nd set: $4 \times 4 = 16$. 3rd set: 64. This is a G.P. with $a=4, r=4$.
Sum Letters
Total letters up to 8th set ($S_8$):
$S_8 = \frac{4(4^8 – 1)}{4 – 1} = \frac{4(65536 – 1)}{3} = \frac{4(65535)}{3} = 87380$.
$S_8 = \frac{4(4^8 – 1)}{4 – 1} = \frac{4(65536 – 1)}{3} = \frac{4(65535)}{3} = 87380$.
Cost
$87380 \times 0.50$ Rupees.
Total Cost: Rs 43,690
Q16
16. Deposited Rs 10000 at 5% simple interest. Find amount in 15th year and after 20 years.
Interest
$I = \frac{10000 \times 5}{100} = 500$ per year.
Sequence
Amounts form an AP: 10000 (Start), 10500 (End Yr 1), 11000 (End Yr 2)…
15th Year
Amount *in* the 15th year is the amount accumulated after 14 years.
$A_{15} = 10000 + 14(500) = 10000 + 7000 = 17000$.
$A_{15} = 10000 + 14(500) = 10000 + 7000 = 17000$.
After 20 Yrs
$A_{20} = 10000 + 20(500) = 10000 + 10000 = 20000$.
15th Year: Rs 17,000
After 20 Years: Rs 20,000
After 20 Years: Rs 20,000
Q17
17. Machine cost Rs 15625. Depreciates 20% each year. Value after 5 years?
Logic
Value becomes $(100-20)\% = 80\%$ of previous year. This is a G.P.
Ratio
$r = \frac{80}{100} = \frac{4}{5}$.
Formula
$V_5 = P \times r^5 = 15625 \times (\frac{4}{5})^5$.
Calculate
$15625 \times \frac{1024}{3125} = 5 \times 1024 = 5120$.
Estimated Value: Rs 5,120
Q18
18. 150 workers. 4 drop out each day. Job took 8 more days than expected. Find total days.
Let
Let work finish in $x$ days if no one dropped out. Total work = $150x$.
Actual
Workers dropped out. Sequence: 150, 146, 142… (AP with $d=-4$).
Time
Actual time taken = $n = x + 8$ days.
Equation
Sum of workers over $n$ days = Total Work.
$\frac{n}{2}[2(150) + (n-1)(-4)] = 150(n-8)$.
$\frac{n}{2}[2(150) + (n-1)(-4)] = 150(n-8)$.
Simplify
$\frac{n}{2}[300 – 4n + 4] = 150n – 1200 \Rightarrow n(152 – 2n) = 150n – 1200$.
Quadratic
$152n – 2n^2 = 150n – 1200 \Rightarrow 2n^2 – 2n – 1200 = 0 \Rightarrow n^2 – n – 600 = 0$.
Solve
$(n-25)(n+24) = 0 \Rightarrow n = 25$ (days cannot be negative).
Work completed in 25 days.