CBSE · NCERT · Class 9 Maths · Chapter 8

NCERT Solutions: Class 9 Maths Chapter 8 - Predicting What Comes Next: Exploring Sequences and Progressions

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Chapter-wise NCERT intext questions and exercise answers for Predicting What Comes Next: Exploring Sequences and Progressions, grounded in the official textbook.

Questions are taken verbatim from the NCERT textbook; answers were grounded against the chapter's content during generation. Items needing review are marked.
Sections in this chapter
Exercise 8.1 6Exercise 8.2 7Exercise 8.3 7Exercise 8.4 15
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1Exercise 8.16 questions
Q.1Find the first five terms of the sequence in which the nth term is given by (i) $t_n = 3n - 4$, (ii) $t_n = 2 - 5n$, and (iii) $t_n = n^2 - 2n + 3$ for $n \geq 1$.v
Solution

Substitute $n=1,2,3,4,5$ in each formula. For example, for $t_n=3n-4$, the terms are $3(1)-4=-1$, $3(2)-4=2$, and so on.

Answer:

(i) $-1, 2, 5, 8, 11$
(ii) $-3, -8, -13, -18, -23$
(iii) $2, 3, 6, 11, 18$

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Q.2Find the 10th and 15th terms of the sequence $t_n = 5n - 3$ for $n \geq 1$.v
Solution

$t_{10}=5(10)-3=47$ and $t_{15}=5(15)-3=72$.

Answer:

$t_{10}=47$ and $t_{15}=72$.

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Q.3Determine whether 97 and 172 are terms of the sequence $t_n = 5n - 3$ for $n \geq 1$.v
Solution

Solve $5n-3=97$: $5n=100$, so $n=20$. Solve $5n-3=172$: $5n=175$, so $n=35$. Both positions are natural numbers, so both are terms.

Answer:

Yes. $97$ is the 20th term and $172$ is the 35th term.

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Q.4Which term of the sequence $t_n = 5n - 3$ for $n \geq 1$ is 607?v
Solution

Set $5n-3=607$. Then $5n=610$, so $n=122$.

Answer:

$607$ is the 122nd term.

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Q.5A sequence is given by the recursive rule $t_1 = -5$, $t_{n+1} = t_n + 3$ for $n \geq 1$. Find the first five terms of the sequence. Is 52 a term of this sequence? If so, which term is it?v
Solution

Starting with $t_1=-5$, add 3 each time to get $-5,-2,1,4,7$. The explicit form is $t_n=-5+3(n-1)$. Set $-5+3(n-1)=52$, so $3(n-1)=57$, $n-1=19$, and $n=20$.

Answer:

The first five terms are $-5,-2,1,4,7$. Yes, $52$ is the 20th term.

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Q.6Let $T_1 = 1$, $T_2 = 2$, $T_3 = 4$, and $T_n = T_{n-1} + T_{n-2} + T_{n-3}$ for $n \geq 4$. Find $T_4$, $T_5$, $T_6$, $T_7$, and $T_8$.v
Solution

$T_4=4+2+1=7$. $T_5=7+4+2=13$. $T_6=13+7+4=24$. $T_7=24+13+7=44$. $T_8=44+24+13=81$.

Answer:

$T_4=7$, $T_5=13$, $T_6=24$, $T_7=44$, $T_8=81$.

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2Exercise 8.27 questions
Q.1Find the 10th and 26th terms of the AP: $3, 8, 13, 18, \ldots$.v
Solution

Here $a=3$ and $d=5$. Using $t_n=a+(n-1)d$, $t_{10}=3+9(5)=48$ and $t_{26}=3+25(5)=128$.

Answer:

$t_{10}=48$ and $t_{26}=128$.

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Q.2Which term of the AP: $21, 18, 15, \ldots$ is $-81$? Also, is 0 a term of this AP? Give reasons for your answer.v
Solution

Here $a=21$ and $d=-3$, so $t_n=21+(n-1)(-3)=24-3n$. For $-81$, solve $24-3n=-81$, giving $n=35$. For 0, solve $24-3n=0$, giving $n=8$. Both are natural-number positions.

Answer:

$-81$ is the 35th term. Yes, $0$ is the 8th term.

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Q.3Find the nth term of the AP: $11, 8, 5, 2, \ldots$. Write the recursive rule for this AP.v
Solution

The first term is $a=11$ and the common difference is $d=-3$. Therefore $t_n=11+(n-1)(-3)=14-3n$. The recursive rule says to start with 11 and subtract 3 each time.

Answer:

$t_n=14-3n$. Recursive rule: $t_1=11$ and $t_n=t_{n-1}-3$ for $n\geq2$.

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Q.4An AP consists of 50 terms in which the 3rd term is 12 and the last term is 106. Find the 29th term. (Hint: If 'a' is the first term and 'd' the common difference, then we arrive at the equations $a + 2d = 12$ and $a + 49d = 106$. Solve this pair of linear equations for 'a' and 'd'.)v
Solution

From $a+2d=12$ and $a+49d=106$, subtract to get $47d=94$, so $d=2$. Then $a+4=12$, so $a=8$. Thus $t_{29}=a+28d=8+56=64$.

Answer:

The 29th term is $64$.

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Q.5How many 2-digit numbers are divisible by 3? What is the sum of all these 2-digit numbers?v
Solution

The two-digit multiples of 3 form the AP $12,15,18,\ldots,99$. The number of terms is $\dfrac{99-12}{3}+1=30$. Sum $=\dfrac{30}{2}(12+99)=15\times111=1665$.

Answer:

There are $30$ such numbers, and their sum is $1665$.

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Q.6Harish started work at an annual salary of `5,00,000 and received an increment of `20,000 each year. After how many years did his income reach `7,00,000?v
Solution

The salaries form an AP with first term `5,00,000 and common difference `20,000. Solve $500000+(n-1)20000=700000$. Then $(n-1)20000=200000$, so $n-1=10$ and $n=11$.

Answer:

His salary reaches `7,00,000 after 10 increments, i.e. in the 11th year of work.

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Q.7A child arranges marbles in rows so that the first row has 1 marble, the second has 2 marbles, the third has 3, and so on up to 25 rows. How many marbles does the child use in all?v
Solution

The total is $1+2+3+\cdots+25=\dfrac{25\times26}{2}=325$.

Answer:

$325$ marbles.

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3Exercise 8.37 questions
Q.1Find the 12th term of a GP with common ratio 2, whose 8th term is 192.v
Solution

In a GP with ratio 2, moving from the 8th term to the 12th term multiplies by $2^4$. Therefore $t_{12}=192\times2^4=192\times16=3072$.

Answer:

The 12th term is $3072$.

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Q.2Find the 10th and nth terms of the GP: $5, 25, 125, \ldots$.v
Solution

The first term is 5 and the common ratio is 5. Thus $t_n=5\cdot5^{n-1}=5^n$. Hence $t_{10}=5^{10}=9765625$.

Answer:

$t_{10}=9765625$ and $t_n=5^n$.

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Q.3A sequence is given by the recursive rule $t_1 = 2$, $t_{n+1} = 3t_n - 2$ for $n \geq 1$. Which term of the sequence is 730?v
Solution

Let $s_n=t_n-1$. Then $s_1=1$ and $s_{n+1}=t_{n+1}-1=3t_n-3=3(t_n-1)=3s_n$. So $s_n=3^{n-1}$ and $t_n=3^{n-1}+1$. Set $3^{n-1}+1=730$, so $3^{n-1}=729=3^6$. Hence $n=7$.

Answer:

$730$ is the 7th term.

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Q.4Which term of the GP: $2, 6, 18, \ldots$ is 4374? Write the explicit formula as well as the recursive formula for the nth term.v
Solution

The common ratio is 3. Solve $2\cdot3^{n-1}=4374$. Then $3^{n-1}=2187=3^7$, so $n=8$.

Answer:

$4374$ is the 8th term. Explicit formula: $t_n=2\cdot3^{n-1}$. Recursive formula: $t_1=2$, $t_n=3t_{n-1}$ for $n\geq2$.

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Q.5A ball is dropped from a height of 80 metres. After hitting the ground, it bounces back to 60% of the height from which it fell. It continues bouncing in this way — each time rising to 60% of the previous height. (i) What height does the ball reach after the 5th bounce? (ii) What is the total vertical distance the ball has travelled by the time it hits the ground for the 6th time?v
Solution

The bounce heights form a GP with first bounce height $80(0.6)=48$ and ratio $0.6$. The 5th bounce height is $80(0.6)^5=6.2208$ m. By the time the ball hits the ground for the 6th time, it has fallen 80 m initially and then gone up and down through the first five bounce heights. Total distance $=80+2(48+28.8+17.28+10.368+6.2208)=301.3376$ m.

Answer:

(i) $6.2208$ m
(ii) $301.3376$ m

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Q.6Which term of the sequence $2, 2\sqrt2, 4, \ldots$ is 128?v
Solution

This is a GP with first term 2 and common ratio $\sqrt2$. Hence $t_n=2(\sqrt2)^{n-1}$. Set $2(\sqrt2)^{n-1}=128$, so $(\sqrt2)^{n-1}=64=(\sqrt2)^{12}$. Therefore $n-1=12$ and $n=13$.

Answer:

$128$ is the 13th term.

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Q.7Fig. 8.12 shows Stages 0 to 3 of the Sierpiński square carpet. Stage 0 of this fractal is a square sheet of paper. To construct Stage 1, each side of the square is trisected and the points of trisection of opposite sides are joined to obtain nine smaller squares. The centre square is then removed and the 8 smaller squares are retained, leaving a square hole in the centre. The same process is repeated on the eight smaller shaded squares to obtain Stage 2 and so on. (i) How many red squares are there in Stages 0 to 3? (ii) Can you predict the number of red squares in Stages 4 and 5? (iii) Can you find a rule for the number of red squares at the nth stage? Write the explicit formula as well as the recursive formula for the number of red squares at any stage. (iv) Suppose the area of the square in Stage 0 is 1 square unit. What is the area of the red region in Stages 1, 2 and 3? What will be the area of the red region in Stages 4 and 5? Find the explicit as well as the recursive formula for the area of the red region at the nth stage. What happens to this area as n, the number of stages, goes on increasing?v
Solution

At each stage, every retained red square produces 8 smaller retained red squares, so the count is multiplied by 8 each time. Each stage also keeps $8$ out of $9$ equal area parts from every retained square, so the total red area is multiplied by $8/9$ at each step.

Answer:

(i) Stages 0 to 3 have $1,8,64,512$ red squares.
(ii) Stages 4 and 5 have $4096$ and $32768$ red squares.
(iii) If Stage 0 is counted as $n=0$, the number of red squares is $R_n=8^n$; recursively, $R_0=1$ and $R_n=8R_{n-1}$ for $n\geq1$.
(iv) Red areas for Stages 1, 2, 3 are $\dfrac89,\dfrac{64}{81},\dfrac{512}{729}$. For Stages 4 and 5 they are $\dfrac{4096}{6561}$ and $\dfrac{32768}{59049}$. Explicitly, $A_n=\left(\dfrac89\right)^n$; recursively, $A_0=1$ and $A_n=\dfrac89A_{n-1}$. As n increases, the area approaches 0.

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4Exercise 8.415 questions
Q.1Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.v
Solution

Let the first term be a and common difference be d. Then $a+10d=38$ and $a+15d=73$. Subtracting gives $5d=35$, so $d=7$. Then $a=38-70=-32$. Hence $t_{31}=a+30d=-32+210=178$.

Answer:

The 31st term is $178$.

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Q.2Determine the AP whose third term is 16 and whose 7th term exceeds the 5th term by 12.v
Solution

Let the AP have first term a and common difference d. The third term is $a+2d=16$. Also, the 7th term exceeds the 5th by 12, so $(a+6d)-(a+4d)=12$, giving $2d=12$ and $d=6$. Then $a+12=16$, so $a=4$.

Answer:

The AP is $4,10,16,22,28,\ldots$.

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Q.3How many three-digit numbers are divisible by 7? (Hint: All three-digit numbers divisible by 7 form an AP. Find the smallest and largest such three-digit numbers.)v
Solution

The smallest three-digit multiple of 7 is $105$ and the largest is $994$. They form an AP with common difference 7. Number of terms $=\dfrac{994-105}{7}+1=127+1=128$.

Answer:

There are $128$ three-digit numbers divisible by 7.

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Q.4How many multiples of 4 lie between 10 and 250? (Hint: All multiples of 4 form an AP. Find the smallest and largest multiples of 4 between 10 and 250.)v
Solution

The first multiple of 4 greater than 10 is 12 and the last multiple of 4 less than 250 is 248. They form an AP with common difference 4. Number of terms $=\dfrac{248-12}{4}+1=59+1=60$.

Answer:

$60$ multiples.

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Q.5Find a GP for which the sum of the first two terms is $-4$ and the fifth term is 4 times the third term.v
Solution

Let the first term be a and common ratio r. The conditions are $a+ar=-4$ and $ar^4=4ar^2$. Assuming non-zero terms, $r^2=4$, so $r=2$ or $r=-2$. If $r=2$, then $3a=-4$, so $a=-4/3$. If $r=-2$, then $a(1-2)=-4$, so $a=4$.

Answer:

Two possible GPs are $-\dfrac{4}{3},-\dfrac{8}{3},-\dfrac{16}{3},\ldots$ and $4,-8,16,-32,64,\ldots$.

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Q.6Find all possible ways of expressing 100 as the sum of consecutive natural numbers.v
Solution

For k consecutive natural numbers starting at a, the sum is $\dfrac{k}{2}(2a+k-1)=100$. Checking positive integer values gives $(k,a)=(1,100),(5,18),(8,9)$. If a single-term sum is excluded, the two non-trivial ways are the 5-term and 8-term sums.

Answer:

$100=100$; $100=18+19+20+21+22$; $100=9+10+11+12+13+14+15+16$.

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Q.7The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of the 2nd hour, 4th hour and nth hour?v
Solution

The population starts at 30 and is multiplied by 2 each hour. After n hours it is $30\cdot2^n$. Thus after 2 hours it is $30\cdot4=120$, and after 4 hours it is $30\cdot16=480$.

Answer:

End of 2nd hour: $120$ bacteria. End of 4th hour: $480$ bacteria. End of nth hour: $30\cdot2^n$ bacteria.

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Q.8The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.v
Solution

Let the AP have first term a and common difference d. Then $t_4+t_8=(a+3d)+(a+7d)=2a+10d=24$, so $a+5d=12$. Also $t_6+t_{10}=(a+5d)+(a+9d)=2a+14d=44$, so $a+7d=22$. Subtracting gives $2d=10$, so $d=5$. Then $a+25=12$, so $a=-13$.

Answer:

The first three terms are $-13,-8,-3$.

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Q.9Find the smallest value of n such that the sum of the first n natural numbers is greater than 1,000.v
Solution

The sum of the first n natural numbers is $\dfrac{n(n+1)}{2}$. For $n=44$, the sum is $\dfrac{44\times45}{2}=990$. For $n=45$, the sum is $\dfrac{45\times46}{2}=1035$, which is greater than 1000. Hence the smallest value is 45.

Answer:

$n=45$.

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Q.10Which term of the GP: $2, 8, 32, \ldots$ is 131072? Write the explicit formula as well as the recursive formula for the nth term.v
Solution

The common ratio is 4. Solve $2\cdot4^{n-1}=131072$. Since $131072=2^{17}$ and $2\cdot4^{n-1}=2^{2n-1}$, we get $2n-1=17$, so $n=9$.

Answer:

$131072$ is the 9th term. Explicit formula: $t_n=2\cdot4^{n-1}$. Recursive formula: $t_1=2$, $t_n=4t_{n-1}$ for $n\geq2$.

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Q.11The sum of the first three terms of a GP is $\dfrac{13}{12}$ and their product is $-1$. Find the common ratio and the terms.v
Solution

Write the three GP terms as $\dfrac{a}{r},a,ar$. Their product is $a^3=-1$, so $a=-1$. Thus the terms are $-\dfrac{1}{r},-1,-r$, and their sum gives $-\dfrac{1}{r}-1-r=\dfrac{13}{12}$. Multiplying by $12r$ gives $12r^2+25r+12=0$, so $r=-\dfrac34$ or $r=-\dfrac43$. These give the two listed GPs.

Answer:

The terms are $\dfrac{4}{3},-1,\dfrac{3}{4}$ with common ratio $-\dfrac{3}{4}$, or $\dfrac{3}{4},-1,\dfrac{4}{3}$ with common ratio $-\dfrac{4}{3}$.

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Q.12If the 4th, 10th and 16th terms of a GP are x, y and z respectively, prove that x, y, z are in GP.v
Answer:

Let the GP have first term a and common ratio r. Then $x=ar^3$, $y=ar^9$ and $z=ar^{15}$. Now $\dfrac{y}{x}=r^6$ and $\dfrac{z}{y}=r^6$. Since the two ratios are equal, x, y and z are in GP.

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Q.13The sum of the first three terms of a geometric progression is 26, and the sum of their squares is 364. Find the terms of the GP.v
Solution

Let the three terms be $\dfrac{b}{r}, b, br$. Their sum is 26 and the sum of their squares is 364. Solving gives middle term $b=6$. Then the first and third terms have sum $20$ and product $36$, so they are roots of $u^2-20u+36=0$, namely $2$ and $18$. Hence the GP is $2,6,18$ or the reverse $18,6,2$.

Answer:

The terms are $2,6,18$ or $18,6,2$.

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Q.14Suppose $P_1 = 1$, $P_2 = 2$ and for $n > 2$, $P_n = P_1 + P_2 + \cdots + P_{n-1} + 1$. Find the values of $P_1, P_2, \ldots, P_8$. Can you find a simpler recursive formula for $P_n$? Can you give an explicit formula?v
Solution

$P_3=1+2+1=4$, $P_4=1+2+4+1=8$, and the pattern continues by doubling. This happens because $P_n=(P_1+\cdots+P_{n-1})+1$, while $P_{n-1}=(P_1+\cdots+P_{n-2})+1$, so $P_n=P_{n-1}+P_{n-1}=2P_{n-1}$.

Answer:

$P_1,\ldots,P_8$ are $1,2,4,8,16,32,64,128$. A simpler recursion is $P_1=1$, $P_n=2P_{n-1}$ for $n\geq2$. The explicit formula is $P_n=2^{n-1}$.

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Q.15Suppose $W_1 = 1$, $W_2 = 2$ and for $n > 2$, $W_n = W_1 + W_2 + \cdots + W_{n-2} + 2$. Find the values of $W_1, W_2, \ldots, W_8$. Do you recognise this sequence?v
Solution

$W_3=W_1+2=3$, $W_4=W_1+W_2+2=5$, $W_5=W_1+W_2+W_3+2=8$, and similarly $W_6=13$, $W_7=21$, $W_8=34$. The terms follow $W_n=W_{n-1}+W_{n-2}$ for $n\geq3$.

Answer:

$W_1,\ldots,W_8$ are $1,2,3,5,8,13,21,34$. This is the Fibonacci/Virahānka sequence starting with 1 and 2.

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