Anand Classes provides NCERT Solutions for Class 11 Maths Chapter 8 – Sequences and Series (Miscellaneous Exercise) with step-by-step solutions and detailed explanations following the latest CBSE and NCERT syllabus. This exercise includes mixed problems from Arithmetic Progression (A.P.), Geometric Progression (G.P.), Harmonic Progression (H.P.), and the relation between A.M., G.M., and H.M. These comprehensive solutions help students develop a deeper understanding of mathematical patterns, series summation, and progression relationships. The notes and solved examples are ideal for CBSE Class 11 students, as well as those preparing for JEE Main, JEE Advanced, NDA, and CUET exams. Click the print button to download study material and notes in PDF format.
Question 1. If $f$ is a function satisfying $f(x + y) = f(x) f(y)$ for all $x, y \in \mathbb{N}$ such that $f(1) = 3$ and $\displaystyle \sum_{x=1}^{x=n} f(x) = 120$, find the value of $n$.
Solution:
We are given that the function satisfies the relation
$$f(x + y) = f(x) \times f(y) \quad \text{for all } x, y \in \mathbb{N} \quad \text{…(1)}$$
and $f(1) = 3$.
Let us find the next few terms of the sequence generated by $f(x)$.
Putting $x = y = 1$ in equation (1),
$$f(2) = f(1 + 1) = f(1) \times f(1) = 3 \times 3 = 9.$$
Next,
$$f(3) = f(2 + 1) = f(2) \times f(1) = 9 \times 3 = 27.$$
Similarly,
$$f(4) = f(3 + 1) = f(3) \times f(1) = 27 \times 3 = 81.$$
Hence, the sequence of $f(x)$ is:
$$3, 9, 27, 81, \ldots$$
This clearly forms a geometric progression (G.P.) with
first term $a = 3$ and common ratio $r = 3$.
We know that the sum of $n$ terms of a G.P. is given by
$$S_n = \frac{a(r^n – 1)}{r – 1}.$$
According to the question,
$$S_n = 120.$$
Substituting the values of $a$ and $r$,
$$120 = \frac{3(3^n – 1)}{3 – 1}.$$
Simplifying,
$$120 = \frac{3(3^n – 1)}{2}$$
$$\Rightarrow 240 = 3(3^n – 1)$$
$$\Rightarrow 80 = 3^n – 1$$
$$\Rightarrow 3^n = 81 = 3^4.$$
Therefore,
$$n = 4.$$
✅ Hence, the number of terms is 4.
Question 2. The sum of some terms of a G.P. is 315 whose first term and common ratio are 5 and 2 respectively. Find the last term and the number of terms.
Solution:
Let the total number of terms be $n$.
We know that the sum of $n$ terms of a G.P. is
$$S_n = \frac{a(r^n – 1)}{r – 1}.$$
Here, $a = 5$, $r = 2$, and $S_n = 315$.
Substituting,
$$315 = \frac{5(2^n – 1)}{2 – 1}.$$
Simplify:
$$315 = 5(2^n – 1)$$
$$\Rightarrow 63 = 2^n – 1$$
$$\Rightarrow 2^n = 64.$$
We know that $64 = 2^6$, therefore
$$n = 6.$$
Now, the last term of the G.P. is given by
$$l = ar^{n – 1}.$$
Substituting,
$$l = 5(2)^{6 – 1} = 5(2^5) = 5 \times 32 = 160.$$
✅ Hence, the last term is 160 and the number of terms is 6.
Question 3. The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of the G.P.
Solution:
Let the first term and common ratio of the G.P. be $a$ and $r$ respectively.
Given that $a = 1$
Now,
the third term of the G.P. is
$$a_3 = ar^2 = r^2 \quad \text{…(1)}$$
and the fifth term is
$$a_5 = ar^4 = r^4 \quad \text{…(2)}$$
According to the question,
$$a_3 + a_5 = 90$$
Substituting from (1) and (2):
$$r^2 + r^4 = 90$$
$$r^4 + r^2 – 90 = 0$$
Let $r^2 = x$.
Then the equation becomes
$$x^2 + x – 90 = 0$$
Solving for $x$ using the quadratic formula:
$$x = \frac{-1 \pm \sqrt{1 + 360}}{2}$$
$$x = \frac{-1 \pm 19}{2}$$
So,
$$x = 9 \quad \text{or} \quad x = -10$$
Since $r^2$ must be positive, we take $r^2 = 9$
Therefore,
$$r = \pm 3$$
Hence, the common ratio of the G.P. is $r = \pm 3$.
Question 4. The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.
Solution:
Let the three numbers in G.P. be $a$, $ar$, and $ar^2$.
According to the question,
$$a + ar + ar^2 = 56$$
or,
$$a(1 + r + r^2) = 56$$
Hence,
$$a = \frac{56}{1 + r + r^2} \quad \text{…(1)}$$
It is also given that
the numbers $(a – 1)$, $(ar – 7)$, and $(ar^2 – 21)$ form an A.P.
For numbers to be in A.P., the difference between consecutive terms must be equal.
Therefore,
$$(ar – 7) – (a – 1) = (ar^2 – 21) – (ar – 7)$$
Simplifying,
$$ar – a – 6 = ar^2 – ar – 14$$
$$ar^2 – 2ar + a = 8$$
$$a(r – 1)^2 = 8 \quad \text{…(2)}$$
Substitute the value of $a$ from equation (1) into equation (2):
$$\frac{56}{1 + r + r^2}(r – 1)^2 = 8$$
Simplify:
$$7(r^2 – 2r + 1) = 1 + r + r^2$$
$$7r^2 – 14r + 7 = r^2 + r + 1$$
$$6r^2 – 15r + 6 = 0$$
Factorizing,
$$(6r – 3)(r – 2) = 0$$
Hence,
$$r = 2 \quad \text{or} \quad r = \frac{1}{2}$$
Case 1: When $r = 2$
Substitute $r = 2$ in equation (1):
$$a = \frac{56}{1 + 2 + 4} = \frac{56}{7} = 8$$
Therefore, the three numbers are
$$a = 8, \quad ar = 16, \quad ar^2 = 32.$$
Case 2: When $r = \frac{1}{2}$
Substitute $r = \frac{1}{2}$ in equation (1):
$$a = \frac{56}{1 + \frac{1}{2} + \frac{1}{4}} = \frac{56}{\frac{7}{4}} = 32$$
Therefore, the three numbers are
$$a = 32, \quad ar = 16, \quad ar^2 = 8.$$
Hence, in both cases, the three numbers in G.P. are 8, 16, and 32.
Question.5 : A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, find its common ratio.
Solution
Let the G.P. have $2n$ terms and first term $a$ and common ratio $r$.
So the terms are
$$a, ar, ar^2, ar^3, \dots, ar^{2n-1}$$
Sum of all $2n$ terms is
$$S_{\text{total}}=a\frac{1-r^{2n}}{1-r}\quad\text{(for }r\neq1\text{)}.$$
Sum of terms occupying odd places is the sum of the $n$ terms
$$a, ar^2, ar^4, \dots, ar^{2(n-1)}$$
This is a G.P. with first term $a$, number of terms becomes $n$ and common ratio $r^2$, so
$$S_{\text{odd}}=a\frac{1-(r^2)^{n}}{1-r^2}.$$
Given $S_{\text{total}}=5 \cdot S_{\text{odd}}$, therefore
$$
a\frac{1-r^{2n}}{1-r}=5 \cdot a\frac{1-r^{2n}}{1-r^2}
$$
If $1-r^{2n}=0$ then all terms are zero and the statement is trivial, so assume $1-r^{2n}\neq0$ and cancel $a(1-r^{2n})$ to get
$$\frac{1}{1-r}=\frac{5}{1-r^2}$$
Since $1-r^2=(1-r)(1+r)$ we have
$$\frac{1}{1-r}=\frac{5}{(1-r)(1+r)}$$
Cancel $1-r$ (note $r\neq1$ or else the given relation fails unless $a=0$) and obtain
$$1=\frac{5}{1+r}$$
Hence
$$1+r=5\quad\Rightarrow\quad r=4$$
We reject $r=1$ because then $S_{\text{total}}=2na$ and $S_{\text{odd}}=na$ would give $2na=5na$, impossible unless $a=0$ which is trivial.
Therefore the common ratio is $\boxed{r=4.}$
Question.6 : If $\dfrac{a+bx}{a-bx}=\dfrac{b+cx}{b-cx}=\dfrac{c+dx}{c-dx}$, show that $a,b,c,d$ are in G.P.
Solution
Let the common value be $k$, so
$$\frac{a+bx}{a-bx}=k,\qquad \frac{b+cx}{b-cx}=k,\qquad \frac{c+dx}{c-dx}=k.$$
From the first equality $\dfrac{a+bx}{a-bx}=\dfrac{b+cx}{b-cx}$ cross multiply:
$$ (a+bx)(b-cx)=(b+cx)(a-bx).$$
Expand both sides:
$$ ab-acx+b^2x-bcx^2 = ab-b^2x+acx-bcx^2.$$
Cancel common terms $ab$ and $-bcx^2$ and collect like terms:
$$ b^2x + b^2x = acx + acx$$
$$ 2b^2x = 2acx.$$
If $x\neq0$ we get
$$ b^2 = ac.$$
Thus
$$\frac{b}{a}=\frac{c}{b}.$$
Similarly equate the second and third ratios:
$$\frac{b+cx}{b-cx}=\frac{c+dx}{c-dx}$$
Cross multiply and expand to obtain (after cancelations)
$$ c^2 = bd.$$
Thus
$$\frac{c}{b}=\frac{d}{c}.$$
From $b^2=ac$ and $c^2=bd$ we have
$$\frac{b}{a}=\frac{c}{b}=\frac{d}{c}.$$
Hence $a,b,c,d$ form a geometric progression with common ratio $\dfrac{b}{a}$.
Therefore
$$\boxed{a,b,c,d\ \text{are in G.P.}}$$
Question.7 : Let $S$ be the sum, $P$ the product, and $R$ the sum of reciprocals of $n$ terms in a G.P. Prove that$$P^2 R^n = S^n$$
Solution :
Suppose the terms of the G.P. are
$a, ar, ar^2, ar^3, \dots, ar^{,n-1}$.
The sum of the terms is
$$
S = \frac{a(r^n – 1)}{r – 1} \quad \text{….. (1)}
$$
The product of the terms is
$$
P = a \cdot ar \cdot ar^2 \cdots ar^{,n-1}
$$
$$
P = a^n \cdot r^{1+2+\dots+(n-1)}
$$
$$
P = a^n r^{,\frac{n(n-1)}{2}}
$$
The sum of reciprocals is
$$
R = \frac{1}{a} + \frac{1}{ar} + \frac{1}{ar^2} + \dots + \frac{1}{ar^{,n-1}}
$$
$$
R = \frac{r^n – 1}{a r^{,n-1}(r – 1)}
$$
Now, compute $P^2 R^n$:
$$
P^2 R^n = \left(a^n r^{,\frac{n(n-1)}{2}}\right)^2 \left(\frac{r^n – 1}{a r^{,n-1}(r – 1)}\right)^n
$$
Split step-by-step:
$$
P^2 R^n = a^{2n} r^{,n(n-1)} \cdot \frac{(r^n – 1)^n}{a^n r^{,n(n-1)} (r – 1)^n}
$$
$$
P^2 R^n = \frac{a^n (r^n – 1)^n}{(r – 1)^n}
$$
$$
P^2 R^n = \left[\frac{a(r^n – 1)}{r – 1}\right]^n
$$
From (1), we get
$$
P^2 R^n = S^n
$$
Hence, proved.
Question.8 : If $a, b, c, d$ are in G.P., prove that $(a^n + b^n), (b^n + c^n), (c^n + d^n)$ are in G.P.
Solution :
Since $a, b, c, d$ are in G.P., we have
$$
b^2 = ac \quad \text{(1)}, \quad c^2 = bd \quad \text{(2)}, \quad ad = bc \quad \text{(3)}
$$
We need to prove that $(a^n + b^n), (b^n + c^n), (c^n + d^n)$ are in G.P., i.e.,
$$
(b^n + c^n)^2 = (a^n + b^n)(c^n + d^n)
$$
Step 1: Expand L.H.S.
$$
(b^n + c^n)^2 = b^{2n} + 2 b^n c^n + c^{2n}
$$
$$
(b^n + c^n)^2 = (b^2)^n + 2 b^n c^n + (c^2)^n
$$
$$
(b^n + c^n)^2 = (ac)^n + 2 b^n c^n + (bd)^n \quad \text{[From (1) and (2)]}
$$
$$
(b^n + c^n)^2 = a^n c^n + b^n c^n + b^n c^n + b^n d^n
$$
$$
(b^n + c^n)^2 = (a^n + b^n)(c^n + d^n) \quad \text{[Using (3)]}
$$
Step 2: Compare with R.H.S.
$$
(b^n + c^n)^2 = (a^n + b^n)(c^n + d^n)
$$
Hence, $(a^n + b^n), (b^n + c^n), (c^n + d^n)$ are in G.P.
Proved.
Question.9 : If $a$ and $b$ are the roots of $x^2 – 3x + p = 0$ and $c, d$ are the roots of $x^2 – 12x + q = 0$, where $a, b, c, d$ form a G.P., prove that $$ (q + p) : (q – p) = 17 : 15$$
Solution :
Let the common ratio of the G.P. be $r$. Then
$$
b = ar, \quad c = ar^2, \quad d = ar^3
$$
Step 1: Roots $a$ and $b$
Sum of roots:
$$
a + b = 3 \implies a + ar = 3 \implies a(1 + r) = 3 \quad \text{(1)}
$$
Product of roots:
$$
ab = p \implies a \cdot ar = p \implies a^2 r = p \quad \text{(2)}
$$
Step 2: Roots $c$ and $d$
Sum of roots:
$$
c + d = 12 \implies ar^2 + ar^3 = 12 \implies ar^2(1 + r) = 12 \quad \text{(3)}
$$
Product of roots:
$$
cd = q \implies ar^2 \cdot ar^3 = q \implies a^2 r^5 = q \quad \text{(4)}
$$
Step 3: Find $r$
Divide (3) by (1):
$$
\frac{ar^2(1+r)}{a(1+r)} = \frac{12}{3} \implies r^2 = 4 \implies r = \pm 2
$$
Step 4: Case $r = 2$
From (1):
$$
a(1 + 2) = 3 \implies a = 1
$$
From (2):
$$
p = 1^2 \cdot 2 = 2
$$
From (4):
$$
q = 1^2 \cdot 2^5 = 32
$$
Check ratio:
$$
\frac{q+p}{q-p} = \frac{32+2}{32-2} = \frac{34}{30} = \frac{17}{15}
$$
Step 5: Case $r = -2$
From (1):
$$
a(1 – 2) = 3 \implies a = -3
$$
From (2):
$$
p = (-3)^2 \cdot (-2) = -18
$$
From (4):
$$
q = (-3)^2 \cdot (-2)^5 = -288
$$
Check ratio:
$$
\frac{q+p}{q-p} = \frac{-288 + (-18)}{-288 – (-18)} = \frac{-306}{-270} = \frac{17}{15}
$$
Hence, proved.
Question.10 : The ratio of the A.M. and G.M. of two positive numbers $a$ and $b$ is $m:n$. Show that $$a:b = (m + \sqrt{m^2 – n^2}) : (m – \sqrt{m^2 – n^2})$$
Solution :
A.M. and G.M. of $a$ and $b$ are
$$
\text{A.M.} = \frac{a+b}{2}, \quad \text{G.M.} = \sqrt{ab}
$$
Given:
$$
\frac{a+b}{2\sqrt{ab}} = \frac{m}{n}
$$
Apply Componendo and Dividendo:
$$
\frac{a+b + 2\sqrt{ab}}{a+b – 2\sqrt{ab}} = \frac{m+n}{m-n}
$$
Rewrite numerator and denominator as squares:
$$
\frac{(\sqrt{a} + \sqrt{b})^2}{(\sqrt{a} – \sqrt{b})^2} = \frac{m+n}{m-n}
$$
Take square roots:
$$
\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} – \sqrt{b}} = \sqrt{\frac{m+n}{m-n}}
$$
Apply Componendo and Dividendo again:
$$
\frac{(\sqrt{a} + \sqrt{b}) + (\sqrt{a} – \sqrt{b})}{(\sqrt{a} + \sqrt{b}) – (\sqrt{a} – \sqrt{b})} = \frac{\sqrt{m+n} + \sqrt{m-n}}{\sqrt{m+n} – \sqrt{m-n}}
$$
$$
\frac{\sqrt{a}}{\sqrt{b}} = \frac{\sqrt{m+n} + \sqrt{m-n}}{\sqrt{m+n} – \sqrt{m-n}}
$$
Square both sides:
$$
\frac{a}{b} = \frac{(\sqrt{m+n} + \sqrt{m-n})^2}{(\sqrt{m+n} – \sqrt{m-n})^2}
$$
Simplify numerator and denominator:
$$
\frac{a}{b} = \frac{(m+n) + (m-n) + 2\sqrt{(m+n)(m-n)}}{(m+n) + (m-n) – 2\sqrt{(m+n)(m-n)}}
$$
$$
\frac{a}{b} = \frac{2m + 2\sqrt{m^2 – n^2}}{2m – 2\sqrt{m^2 – n^2}}
$$
$$
\frac{a}{b} = \frac{m + \sqrt{m^2 – n^2}}{m – \sqrt{m^2 – n^2}}
$$
Hence, proved.
Question.11 : Find the sum of the following series up to $n$ terms:
(i) $5 + 55 + 555 + \dots$
(ii) $0.6 + 0.66 + 0.666 + \dots$
Solution :
(i) $5 + 55 + 555 + \dots$
Let
$$
S_n = 5 + 55 + 555 + \dots \text{ up to } n \text{ terms.}
$$
Multiply and divide by 9:
$$
S_n = \frac{5}{9} \left[9 + 99 + 999 + \dots \text{ up to } n \text{ terms}\right]
$$
Rewrite each term as $(10^k – 1)$:
$$
S_n = \frac{5}{9} \left[(10-1) + (10^2 – 1) + (10^3 – 1) + \dots + (10^n – 1)\right]
$$
Split the sum:
$$
S_n = \frac{5}{9} \left[(10 + 10^2 + 10^3 + \dots + 10^n) – (1 + 1 + \dots + 1)\right]
$$
Use the formula for geometric series:
$$
10 + 10^2 + \dots + 10^n = \frac{10(10^n – 1)}{10 – 1} = \frac{10(10^n – 1)}{9}
$$
Number of terms = $n$, so sum of $1+1+\dots+n = n$.
Therefore:
$$
S_n = \frac{5}{9} \left[\frac{10(10^n – 1)}{9} – n\right]
$$
Simplify:
$$
S_n = \frac{50(10^n – 1)}{81} – \frac{5n}{9}
$$
(ii) $0.6 + 0.66 + 0.666 + \dots$
Let
$$
S_n = 0.6 + 0.66 + 0.666 + \dots \text{ up to } n \text{ terms.}
$$
Factor 6:
$$
S_n = 6 \left[0.1 + 0.11 + 0.111 + \dots \text{ up to } n \text{ terms}\right]
$$
Multiply and divide by 9:
$$
S_n = \frac{6}{9} \left[0.9 + 0.99 + 0.999 + \dots \right]
$$
$$
S_n = \frac{2}{3} \left[0.9 + 0.99 + 0.999 + \dots \right]
$$
Rewrite each term as $(1 – \frac{1}{10^k})$:
$$
S_n = \frac{2}{3} \left[\left(1 – \frac{1}{10}\right) + \left(1 – \frac{1}{10^2}\right) + \dots + \left(1 – \frac{1}{10^n}\right) \right]
$$
Split the sum:
$$
S_n = \frac{2}{3} \left[n – \frac{1}{10}\left(1 + \frac{1}{10} + \frac{1}{10^2} + \dots + \frac{1}{10^{n-1}}\right) \right]
$$
Use the geometric series formula:
$$
1 + \frac{1}{10} + \frac{1}{10^2} + \dots + \frac{1}{10^{n-1}} = \frac{1 – (1/10)^n}{1 – 1/10} = \frac{1 – 10^{-n}}{9/10} = \frac{10(1 – 10^{-n})}{9}
$$
Substitute:
$$
S_n = \frac{2}{3} \left[n – \frac{1}{10} \cdot \frac{10(1 – 10^{-n})}{9} \right] = \frac{2}{3} \left[n – \frac{1 – 10^{-n}}{9} \right]
$$
Simplify:
$$
S_n = \frac{2n}{3} – \frac{2(1 – 10^{-n})}{27}
$$
Question.12 : Find the 20th term of the series $$
2 \times 4 + 4 \times 6 + 6 \times 8 + \dots $$
Solution :
The $n$th term of the series is
$$
a_n = 2n \times (2n + 2)
$$
Simplify:
$$
a_n = 4n^2 + 4n
$$
For $n = 20$:
$$
a_{20} = 4(20)^2 + 4(20) = 1600 + 80 = 1680
$$
Therefore, the 20th term of the series is $1680$.
Question.13 : A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to pay the balance in annual installments of Rs 500 plus 12% interest on the unpaid amount. Find the total cost of the tractor.
Solution :
Cash paid = Rs 6000
Remaining unpaid amount = Rs 12000 – Rs 6000 = Rs 6000
Total interest is:
$$
\text{Interest} = 12\% \text{ of } 6000 + 12\% \text{ of } 5500 + 12\% \text{ of } 5000 + \dots + 12\% \text{ of } 500
$$
The series $6000, 5500, 5000, \dots, 500$ is an A.P. with first term $a = 6000$ and common difference $d = -500$.
Let the number of terms be $n$.
The $n$th term formula:
$$
a_n = a + (n-1)d
$$
Set $a_n = 500$:
$$
500 = 6000 + (n-1)(-500)
$$
$$
500 = 6000 – 500(n-1)
$$
$$
500(n-1) = 6000 – 500 = 5500
$$
$$
n – 1 = 11 \implies n = 12
$$
Sum of the A.P.:
$$
S_n = \frac{n}{2} [2a + (n-1)d] = \frac{12}{2} [2(6000) + 11(-500)]
$$
$$
S_n = 6 [12000 – 5500] = 6 \cdot 6500 = 39000
$$
Total interest = $12%$ of 39000:
$$
\text{Interest} = 0.12 \times 39000 = 4680
$$
Total cost of tractor = Rs 12000 + Rs 4680 = Rs 16680
Question.14 : Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the balance in annual installments of Rs 1000 plus 10% interest on the unpaid amount. Find the total cost of the scooter.
Solution :
Cash paid = Rs 4000
Remaining unpaid amount = Rs 22000 – Rs 4000 = Rs 18000
Total interest:
$
\text{Interest} = 10\% \text{ of } 18000 + 10\% \text{ of } 17000 + 10\% \text{ of } 16000 + \dots + 10\% \text{ of } 1000
$
The series $18000, 17000, 16000, \dots, 1000$ is an A.P. with first term $a = 18000$ and common difference $d = -1000$.
Let the number of terms be $n$.
$$
1000 = 18000 + (n-1)(-1000)
$$
$$
1000 = 18000 – 1000(n-1)
$$
$$
1000(n-1) = 18000 – 1000 = 17000
$$
$$
n – 1 = 17 \implies n = 18
$$
Sum of the A.P.:
$$
S_n = \frac{n}{2} [2a + (n-1)d] = \frac{18}{2} [2(18000) + 17(-1000)]
$$
$$
S_n = 9 [36000 – 17000] = 9 \cdot 19000 = 171000
$$
Total interest = $10\%$ of 171000:
$$
\text{Interest} = 0.10 \times 171000 = 17100
$$
Total cost of scooter = Rs 22000 + Rs 17100 = Rs 39100
Question.15 : A person writes a letter to four of his friends. Each recipient copies the letter and mails it to four different persons, continuing the chain. If it costs 50 paisa to mail one letter, find the total postage when the 8th set of letters is mailed.
Solution :
The number of letters mailed forms a G.P.:
$$
4, 4^2, 4^3, \dots, 4^8
$$
First term $a = 4$, common ratio $r = 4$, number of terms $n = 8$.
Sum of $n$ terms of a G.P.:
$$
S_n = \frac{a(r^n – 1)}{r – 1} = \frac{4(4^8 – 1)}{4 – 1}
$$
$$
S_n = \frac{4}{3} (65536 – 1) = \frac{4}{3} \cdot 65535 = 87380
$$
Cost to mail one letter = 50 paisa = Rs 0.50
Total cost:
$$
\text{Cost} = 87380 \times 0.50 = \text{Rs } 43690
$$
The amount spent on postage for the 8th set of letters is Rs 43690.
Question.16 : A man deposited Rs 10000 in a bank at 5% simple interest annually. Find the amount in the 15th year and the total amount after 20 years.
Solution :
Principal $P = 10000$, rate $R = 5\%$ per annum, simple interest.
Interest for one year:
$$
I = \frac{5}{100} \times 10000 = 500
$$
Amount in the 15th year:
Simple interest for 14 years (since the first year only earns principal interest):
$$
A_{15} = 10000 + 14 \times 500 = 10000 + 7000 = 17000
$$
Total amount after 20 years:
$$
A_{20} = 10000 + 20 \times 500 = 10000 + 10000 = 20000
$$
Amount in the 15th year is Rs 17000 and total amount after 20 years is Rs 20000.
Question.17 : A manufacturer reckons that the value of a machine, which costs him Rs 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years.
Solution :
Cost of the machine = Rs 15625
Depreciation = 20% per year → remaining value each year = 80% = $\frac{4}{5}$
Value at the end of 5 years:
$$
V_5 = 15625 \times \left(\frac{4}{5}\right)^5
$$
Compute step by step:
$$
\left(\frac{4}{5}\right)^5 = \frac{4^5}{5^5} = \frac{1024}{3125}
$$
$$
V_5 = 15625 \times \frac{1024}{3125} = 5 \times 1024 = 5120
$$
The estimated value of the machine at the end of 5 years is Rs 5120.
Question.18 : 150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on the second day, 4 more on the third day, and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.
Solution :
Let $x$ be the number of days in which 150 workers would finish the work without any dropouts.
Total work = 150 × $x$ worker-days
Number of workers per day forms an A.P.:
$$
150, 146, 142, \dots \text{ (for } x + 8 \text{ days)}
$$
First term $a = 150$, common difference $d = -4$, number of terms $n = x + 8$
Sum of this series:
$$
\text{Work done} = \frac{n}{2} [2a + (n-1)d] = 150 x
$$
Substitute:
$$
150x = \frac{x+8}{2} [2 \cdot 150 + (x+8 -1)(-4)]
$$
$$
150x = \frac{x+8}{2} [300 – 4(x+7)]
$$
$$
150x = \frac{x+8}{2} [300 -4x -28] = \frac{x+8}{2} (272 – 4x)
$$
Multiply both sides by 2:
$$
300x = (x+8)(272 – 4x)
$$
Expand:
$$
300x = 272x – 4x^2 + 2176 – 32x
$$
$$
300x = -4x^2 + 240x + 2176
$$
Bring all terms to one side:
$$
4x^2 + 60x – 2176 = 0
$$
Divide by 4:
$$
x^2 + 15x – 544 = 0
$$
Factor:
$$
(x + 32)(x – 17) = 0
$$
Ignore negative solution: $x = 17$
Including the extra 8 days:
$$
\text{Total days} = 17 + 8 = 25
$$
The work was completed in 25 days.
