NCERT Class 11 Mathematics — Sequences and Series, Miscellaneous Exercise on Chapter 8. All 18 questions solved.

Three kinds of question appear here.

• GP algebra (1 to 10): identities and proofs, all resting on $a_n = ar^{n-1}$, $b^2 = ac$ for three terms in GP, and $S_n = \dfrac{a(r^n-1)}{r-1}$.
• Series sums (11, 12): sequences that are neither AP nor GP, but become manageable after a rewrite.
• Financial and real-world problems (13 to 18): each is an AP or GP in disguise, and identifying which is most of the work.

Key insight. In the instalment problems (13 and 14), the payments are equal but the interest shrinks each year, because it is charged on a balance that falls by a fixed amount. A quantity decreasing by a constant difference is an arithmetic progression — so the total interest is an AP sum, not a GP one. Confusing this with compound interest, which is geometric, is the single most common error in this exercise.

Question 1

If $f$ is a function satisfying $f(x+y) = f(x)f(y)$ for all $x, y \in \mathbf{N}$ such that $f(1) = 3$ and $\displaystyle\sum_{x=1}^{n} f(x) = 120$, find the value of $n$.

Solution. The functional equation forces $f$ to be exponential. Taking $y = 1$ repeatedly:

$$f(2) = f(1)f(1) = 3^2, \qquad f(3) = f(2)f(1) = 3^3, \qquad \ldots, \qquad f(x) = 3^x$$

So the sum is a GP with $a = 3$, $r = 3$:

$$\sum_{x=1}^n 3^x = \frac{3(3^n – 1)}{3-1} = \frac32\left(3^n – 1\right) = 120$$

$$3^n – 1 = 80 \quad\Longrightarrow\quad 3^n = 81 \quad\Longrightarrow\quad n = 4$$

Question 2

The sum of some terms of a GP is $315$, whose first term and common ratio are $5$ and $2$ respectively. Find the last term and the number of terms.

Solution.

$$S_n = \frac{5(2^n – 1)}{2 – 1} = 5(2^n – 1) = 315$$

$$2^n – 1 = 63 \quad\Longrightarrow\quad 2^n = 64 \quad\Longrightarrow\quad n = 6$$

The last term is the $6$th:

$$a_6 = 5 \times 2^5 = 160$$

Question 3

The first term of a GP is $1$. The sum of the third term and fifth term is $90$. Find the common ratio of the GP.

Solution. With $a = 1$:

$$a_3 = r^2, \qquad a_5 = r^4$$

$$r^2 + r^4 = 90$$

Substituting $u = r^2$ gives a quadratic:

$$u^2 + u – 90 = 0 \quad\Longrightarrow\quad (u + 10)(u – 9) = 0$$

so $u = 9$ or $u = -10$. Since $u = r^2 \ge 0$ for real $r$, discard $-10$:

$$r^2 = 9 \quad\Longrightarrow\quad r = \pm3$$

Question 4

The sum of three numbers in GP 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 numbers be $a$, $ar$, $ar^2$. The first condition is

$$a\left(1 + r + r^2\right) = 56 \qquad (1)$$

After the subtractions the three numbers $a – 1$, $ar – 7$, $ar^2 – 21$ are in AP, so the middle is the average of the outer two — equivalently, the two differences are equal:

$$(ar – 7) – (a – 1) = \left(ar^2 – 21\right) – (ar – 7)$$

$$ar – a – 6 = ar^2 – ar – 14$$

$$2ar – a – ar^2 = -8 \quad\Longrightarrow\quad -a\left(r^2 – 2r + 1\right) = -8 \quad\Longrightarrow\quad a(r-1)^2 = 8 \qquad (2)$$

Dividing (1) by (2) eliminates $a$:

$$\frac{1 + r + r^2}{(r-1)^2} = 7 \quad\Longrightarrow\quad 1 + r + r^2 = 7r^2 – 14r + 7$$

$$6r^2 – 15r + 6 = 0 \quad\Longrightarrow\quad 2r^2 – 5r + 2 = 0 \quad\Longrightarrow\quad (2r – 1)(r – 2) = 0$$

so $r = 2$ or $r = \tfrac12$.

For $r = 2$: from (2), $a(1)^2 = 8$, so $a = 8$ and the numbers are $8, 16, 32$.

For $r = \tfrac12$: $a\left(\tfrac12\right)^2 = 8$ gives $a = 32$, and the numbers are $32, 16, 8$ — the same three in reverse.

Check: $8 + 16 + 32 = 56$ ✓, and $7, 9, 11$ is an AP ✓

Question 5

A GP consists of an even number of terms. If the sum of all the terms is $5$ times the sum of terms occupying odd places, then find its common ratio.

Solution. Let there be $2n$ terms:

$$a,\ ar,\ ar^2,\ \ldots,\ ar^{2n-1}$$

All terms:

$$S = \frac{a\left(r^{2n} – 1\right)}{r – 1}$$

Odd places are the $1$st, $3$rd, $5$th, …, that is $a, ar^2, ar^4, \ldots$ — a GP of $n$ terms with ratio $r^2$:

$$S_{\text{odd}} = \frac{a\left((r^2)^n – 1\right)}{r^2 – 1} = \frac{a\left(r^{2n} – 1\right)}{r^2 – 1}$$

The condition $S = 5\,S_{\text{odd}}$ gives

$$\frac{a(r^{2n}-1)}{r-1} = 5 \cdot \frac{a(r^{2n}-1)}{r^2-1}$$

Cancelling the common factor $a\left(r^{2n}-1\right)$ and using $r^2 – 1 = (r-1)(r+1)$:

$$\frac{1}{r-1} = \frac{5}{(r-1)(r+1)} \quad\Longrightarrow\quad r + 1 = 5 \quad\Longrightarrow\quad r = 4$$

Question 6

If $\dfrac{a + bx}{a – bx} = \dfrac{b + cx}{b – cx} = \dfrac{c + dx}{c – dx}$ $(x \ne 0)$, then show that $a$, $b$, $c$ and $d$ are in GP.

Solution. Apply componendo and dividendo to the first equality — that is, add and subtract $1$ on each side and divide:

$$\frac{(a+bx) + (a – bx)}{(a+bx) – (a-bx)} = \frac{(b+cx) + (b-cx)}{(b+cx) – (b-cx)}$$

$$\frac{2a}{2bx} = \frac{2b}{2cx} \quad\Longrightarrow\quad \frac ab = \frac bc$$

Applying the same operation to the second equality gives

$$\frac bc = \frac cd$$

Combining,

$$\frac ab = \frac bc = \frac cd$$

which says exactly that consecutive terms share a common ratio. Hence $a$, $b$, $c$, $d$ are in GP. $\blacksquare$

Question 7

Let $\mathrm S$ be the sum, $\mathrm P$ the product and $\mathrm R$ the sum of reciprocals of $n$ terms in a GP. Prove that $\mathrm P^2\mathrm R^n = \mathrm S^n$.

Solution. Take the GP as $a, ar, \ldots, ar^{n-1}$.

Sum:

$$\mathrm S = \frac{a(r^n – 1)}{r-1}$$

Product:

$$\mathrm P = a^n r^{1+2+\cdots+(n-1)} = a^n r^{\frac{n(n-1)}{2}}$$

Sum of reciprocals: these form a GP with first term $\dfrac1a$ and ratio $\dfrac1r$:

$$\mathrm R = \frac{\frac1a\left(1 – \frac{1}{r^n}\right)}{1 – \frac1r} = \frac1a \cdot \frac{\frac{r^n – 1}{r^n}}{\frac{r-1}{r}} = \frac{r^n – 1}{a\,r^{n-1}(r-1)}$$

Notice that

$$\mathrm R = \frac{1}{a^2 r^{n-1}} \cdot \frac{a(r^n-1)}{r-1} = \frac{\mathrm S}{a^2 r^{n-1}}$$

Therefore

$$\mathrm R^n = \frac{\mathrm S^n}{a^{2n}r^{n(n-1)}}$$

and since $\mathrm P^2 = a^{2n}r^{n(n-1)}$,

$$\mathrm P^2 \mathrm R^n = a^{2n}r^{n(n-1)} \cdot \frac{\mathrm S^n}{a^{2n}r^{n(n-1)}} = \mathrm S^n \qquad \blacksquare$$

Question 8

If $a$, $b$, $c$, $d$ are in GP, prove that $(a^n + b^n)$, $(b^n + c^n)$, $(c^n + d^n)$ are in GP.

Solution. Write $b = ar$, $c = ar^2$, $d = ar^3$. Then

$$a^n + b^n = a^n\left(1 + r^n\right)$$ $$b^n + c^n = a^nr^n\left(1 + r^n\right)$$ $$c^n + d^n = a^nr^{2n}\left(1 + r^n\right)$$

Each is the previous one multiplied by $r^n$, so the three are in GP with common ratio $r^n$. Equivalently, checking the defining condition:

$$\left(b^n + c^n\right)^2 = a^{2n}r^{2n}\left(1+r^n\right)^2 = \left(a^n + b^n\right)\left(c^n + d^n\right) \qquad \blacksquare$$

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 GP, prove that $(q+p) : (q-p) = 17 : 15$.

Solution. From the two quadratics, by the sum and product of roots:

$$a + b = 3, \qquad ab = p, \qquad c + d = 12, \qquad cd = q$$

Let the GP be $a, ar, ar^2, ar^3$, so $b = ar$, $c = ar^2$, $d = ar^3$. Then

$$a + ar = 3 \quad\Longrightarrow\quad a(1 + r) = 3$$ $$ar^2 + ar^3 = 12 \quad\Longrightarrow\quad ar^2(1 + r) = 12$$

Dividing the second by the first:

$$r^2 = 4 \quad\Longrightarrow\quad r = \pm2$$

Take $r = 2$: then $3a = 3$, so $a = 1$ and

$$p = ab = a \cdot ar = a^2r = 2, \qquad q = cd = ar^2 \cdot ar^3 = a^2r^5 = 32$$

$$\frac{q + p}{q – p} = \frac{34}{30} = \frac{17}{15}$$

Take $r = -2$: then $a(-1) = 3$, so $a = -3$ and

$$p = a^2r = 9(-2) = -18, \qquad q = a^2r^5 = 9(-32) = -288$$

$$\frac{q+p}{q-p} = \frac{-306}{-270} = \frac{17}{15}$$

Either way,

$$(q+p) : (q-p) = 17 : 15 \qquad \blacksquare$$

Question 10

The ratio of the AM and GM of two positive numbers $a$ and $b$ is $m : n$. Show that

$$a : b = \left(m + \sqrt{m^2 – n^2}\right) : \left(m – \sqrt{m^2 – n^2}\right)$$

Solution. The hypothesis is

$$\frac{\frac{a+b}{2}}{\sqrt{ab}} = \frac mn \quad\Longrightarrow\quad \frac{a+b}{2\sqrt{ab}} = \frac mn \quad\Longrightarrow\quad a + b = \frac{2m}{n}\sqrt{ab}$$

Setting $k = \dfrac{2m}{n}$, this is the situation of Exercise 8.2 question 28 with $6$ replaced by $k$. Adding and subtracting $2\sqrt{ab}$:

$$\left(\sqrt a + \sqrt b\right)^2 = (k+2)\sqrt{ab}, \qquad \left(\sqrt a – \sqrt b\right)^2 = (k-2)\sqrt{ab}$$

Dividing:

$$\frac{\sqrt a + \sqrt b}{\sqrt a – \sqrt b} = \sqrt{\frac{k+2}{k-2}} = \sqrt{\frac{2m/n + 2}{2m/n – 2}} = \sqrt{\frac{m+n}{m-n}}$$

By componendo and dividendo,

$$\frac{\sqrt a}{\sqrt b} = \frac{\sqrt{m+n} + \sqrt{m-n}}{\sqrt{m+n} – \sqrt{m-n}}$$

Squaring, and rationalising by multiplying above and below by $\left(\sqrt{m+n} + \sqrt{m-n}\right)^2$:

$$\frac ab = \frac{\left(\sqrt{m+n}+\sqrt{m-n}\right)^2}{\left(\sqrt{m+n}\right)^2 – \left(\sqrt{m-n}\right)^2} \cdot \left(\ldots\right)$$

More directly, from $\left(\sqrt{m+n} \pm \sqrt{m-n}\right)^2 = 2m \pm 2\sqrt{m^2-n^2}$:

$$\frac ab = \frac{2m + 2\sqrt{m^2-n^2}}{2m – 2\sqrt{m^2 – n^2}} = \frac{m + \sqrt{m^2-n^2}}{m – \sqrt{m^2-n^2}} \qquad \blacksquare$$

Question 11

Find the sum of the following series up to $n$ terms:

(i) $5 + 55 + 555 + \cdots$    (ii) $0.6 + 0.66 + 0.666 + \cdots$

Solution.

(i) Take out the $5$ and use the $\tfrac19$ trick, as in Exercise 8.2 question 18:

$$5 + 55 + 555 + \cdots = \frac59\left(9 + 99 + 999 + \cdots\right) = \frac59\left[\left(10 – 1\right) + \left(10^2 – 1\right) + \cdots\right]$$

$$= \frac59\left[\frac{10\left(10^n – 1\right)}{9} – n\right] = \frac{50}{81}\left(10^n – 1\right) – \frac{5n}{9}$$

(ii) Now the digits are on the other side of the point:

$$0.6 + 0.66 + 0.666 + \cdots = \frac69\left(0.9 + 0.99 + 0.999 + \cdots\right)$$

$$= \frac23\left[\left(1 – 10^{-1}\right) + \left(1 – 10^{-2}\right) + \cdots + \left(1 – 10^{-n}\right)\right]$$

$$= \frac23\left[n – \left(10^{-1} + 10^{-2} + \cdots + 10^{-n}\right)\right]$$

The inner sum is a GP with $a = \tfrac{1}{10}$ and $r = \tfrac{1}{10}$:

$$10^{-1} + \cdots + 10^{-n} = \frac{\frac{1}{10}\left(1 – 10^{-n}\right)}{1 – \frac{1}{10}} = \frac19\left(1 – 10^{-n}\right)$$

Therefore

$$= \frac{2n}{3} – \frac{2}{27}\left(1 – 10^{-n}\right)$$

Question 12

Find the $20$th term of the series $2 \times 4 + 4 \times 6 + 6 \times 8 + \cdots + n$ terms.

Solution. Look at the pattern of factors: the first factors are $2, 4, 6, \ldots$ and the second are $4, 6, 8, \ldots$ — each an AP. The $n$th term is therefore

$$a_n = (2n)(2n + 2) = 4n(n+1)$$

Checking: $a_1 = 4(1)(2) = 8 = 2 \times 4$ ✓

$$a_{20} = 4 \times 20 \times 21 = 1680$$

Question 13

A farmer buys a used tractor for Rs $12000$. He pays Rs $6000$ cash and agrees to pay the balance in annual instalments of Rs $500$ plus $12\%$ interest on the unpaid amount. How much will the tractor cost him?

Solution. The unpaid balance is $12000 – 6000 = $ Rs $6000$, cleared at Rs $500$ a year, so there are

$$\frac{6000}{500} = 12 \text{ instalments}$$

The interest each year is $12\%$ of the balance outstanding at the start of that year, and that balance falls by Rs $500$ annually:

$$12\% \text{ of } 6000,\ 12\% \text{ of } 5500,\ \ldots,\ 12\% \text{ of } 500$$

$$= 720,\ 660,\ 600,\ \ldots,\ 60$$

This is an AP with $12$ terms, first term $720$, last term $60$. Its sum is

$$S_{12} = \frac{12}{2}(720 + 60) = 6 \times 780 = 4680$$

Total cost:

$$12000 + 4680 = \text{Rs } 16680$$

Note the interest is simple on a falling balance, which makes the yearly amounts an AP — not compound, which would make them a GP.

Question 14

Shamshad Ali buys a scooter for Rs $22000$. He pays Rs $4000$ cash and agrees to pay the balance in annual instalments of Rs $1000$ plus $10\%$ interest on the unpaid amount. How much will the scooter cost him?

Solution. The balance is $22000 – 4000 = $ Rs $18000$, paid off at Rs $1000$ a year, so $18$ instalments.

The annual interest amounts are $10\%$ of $18000, 17000, \ldots, 1000$:

$$1800,\ 1700,\ 1600,\ \ldots,\ 100$$

An AP of $18$ terms from $1800$ down to $100$:

$$S_{18} = \frac{18}{2}(1800 + 100) = 9 \times 1900 = 17100$$

Total cost:

$$22000 + 17100 = \text{Rs } 39100$$

Question 15

A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail it to four different persons with the instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs $50$ paise to mail one letter, find the amount spent on the postage when the $8$th set of letters is mailed.

Solution. The numbers of letters mailed in successive rounds are

$$4,\ 4^2,\ 4^3,\ \ldots,\ 4^8$$

a GP with $a = 4$, $r = 4$, $n = 8$. The total number of letters is

$$S_8 = \frac{4\left(4^8 – 1\right)}{4-1} = \frac{4(65536 – 1)}{3} = \frac{4 \times 65535}{3} = 87380$$

Each costs $50$ paise $= $ Rs $0.50$, so the postage is

$$87380 \times 0.50 = \text{Rs } 43690$$

Question 16

A man deposited Rs $10000$ in a bank at the rate of $5\%$ simple interest annually. Find the amount in the $15$th year since he deposited the amount, and also calculate the total amount after $20$ years.

Solution. Simple interest adds a fixed Rs $500$ each year ($5\%$ of $10000$), so the amounts form an arithmetic progression:

$$10000,\ 10500,\ 11000,\ \ldots$$

with $a = 10000$ and $d = 500$.

Amount in the $15$th year — that is, after $14$ further years of interest:

$$a_{15} = 10000 + 14(500) = 10000 + 7000 = \text{Rs } 17000$$

Total after $20$ years:

$$10000 + 20(500) = 10000 + 10000 = \text{Rs } 20000$$

The word simple is what makes this an AP; compound interest would give a GP with ratio $1.05$.

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. Losing $20\%$ each year means retaining $80\%$, so the values form a GP with ratio $0.8$:

$$15625,\ 15625(0.8),\ 15625(0.8)^2,\ \ldots$$

After $5$ years:

$$15625 \times (0.8)^5 = 15625 \times \frac{4^5}{5^5} = 15625 \times \frac{1024}{3125} = 5 \times 1024 = \text{Rs } 5120$$

Writing $0.8$ as $\tfrac45$ makes the arithmetic exact — $15625 = 5^6$ cancels against $5^5$ neatly.

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 workers dropped out 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 the work have been planned for $n$ days with all $150$ workers, so the total work is

$$150n \text{ worker-days}$$

In practice the job ran for $n + 8$ days, with the workforce falling by $4$ each day:

$$150,\ 146,\ 142,\ \ldots$$

an AP with $a = 150$, $d = -4$, over $n + 8$ terms. The work actually done is

$$S_{n+8} = \frac{n+8}{2}\left[2(150) + (n + 8 – 1)(-4)\right] = \frac{n+8}{2}\left[300 – 4n – 28\right] = \frac{n+8}{2}(272 – 4n)$$

$$= (n+8)(136 – 2n)$$

Equating the two expressions for the same job:

$$(n+8)(136 – 2n) = 150n$$

$$136n – 2n^2 + 1088 – 16n = 150n$$

$$-2n^2 + 120n + 1088 = 150n$$

$$2n^2 + 30n – 1088 = 0 \quad\Longrightarrow\quad n^2 + 15n – 544 = 0$$

$$(n + 32)(n – 17) = 0$$

Rejecting the negative root, $n = 17$. The work was actually completed in

$$n + 8 = 25 \text{ days}$$

Common mistakes

• Question 1, trying to find $f$ by inspection. Setting $y = 1$ turns the functional equation into $f(x+1) = 3f(x)$, which is a GP recursion — that is the whole derivation.
• Question 3, keeping the root $r^2 = -10$. A square cannot be negative for real $r$, so only $r^2 = 9$ survives, giving $r = \pm 3$.
• Question 4, using $\tfrac ar, a, ar$ here. The symmetric form is the right choice when the product is given; here the sum is given and the AP condition is written in terms of consecutive differences, so $a, ar, ar^2$ is simpler.
• Question 5, taking the odd-place terms to have ratio $r$. They are $a, ar^2, ar^4, \ldots$, so their ratio is $r^2$ and there are $n$ of them, not $2n$.
• Questions 13 and 14, treating the interest as compound. The balance falls by a fixed instalment, so the interest amounts form an AP. Summing them as a GP gives a quite different — and wrong — total.
• Question 13, miscounting the instalments. Rs $6000$ at Rs $500$ a year is $12$ payments, and the interest is charged on $6000$ down to $500$, which is also $12$ terms.
• Question 16, applying compound interest. The question says simple, so the yearly addition is a constant Rs $500$ and the sequence is arithmetic.
• Question 17, subtracting $20\%$ five times from the original value. Depreciation applies to the current value each year, so the factor $0.8$ is applied repeatedly — a GP, not a fixed deduction.
• Question 18, forgetting the workforce falls on the second day, not the first. Day 1 has all $150$ workers; the AP is $150, 146, 142, \ldots$ and runs for $n + 8$ terms.

Practise next

• Exercise 9.1 — the next chapter, Straight Lines, which begins a new topic.
• Exercise 8.2 — worth working again alongside questions 4 to 10, since every one of them is a GP identity of the kind practised there.