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

Exercise 8.1

Q1. Find the first five terms of the sequence in which the nth term is given by:

(i) $t_n = 3n – 4$tn​=3n−4
(ii) $t_n = 2 – 5n$tn​=2−5n
(iii) $t_n = n^2 – 2n + 3$tn​=n2−2n+3, for $n \geq 1$n≥1

Solution:

We find terms by putting $n = 1, 2, 3, 4, 5$n=1,2,3,4,5

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(i) tn=3n−4t_n = 3n – 4tn​=3n−4

$t_1 = 3(1) – 4 = -1$t1​=3(1)−4=−1
$t_2 = 3(2) – 4 = 2$t2​=3(2)−4=2
$t_3 = 3(3) – 4 = 5$t3​=3(3)−4=5
$t_4 = 3(4) – 4 = 8$t4​=3(4)−4=8
$t_5 = 3(5) – 4 = 11$t5​=3(5)−4=11

Answer: $-1, 2, 5, 8, 11$−1,2,5,8,11

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(ii) tn=2−5nt_n = 2 – 5ntn​=2−5n

$t_1 = 2 – 5 = -3$t1​=2−5=−3
$t_2 = 2 – 10 = -8$t2​=2−10=−8
$t_3 = 2 – 15 = -13$t3​=2−15=−13
$t_4 = 2 – 20 = -18$t4​=2−20=−18
$t_5 = 2 – 25 = -23$t5​=2−25=−23

Answer: $-3, -8, -13, -18, -23$−3,−8,−13,−18,−23

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(iii) tn=n2−2n+3t_n = n^2 – 2n + 3tn​=n2−2n+3

$t_1 = 1 – 2 + 3 = 2$t1​=1−2+3=2
$t_2 = 4 – 4 + 3 = 3$t2​=4−4+3=3
$t_3 = 9 – 6 + 3 = 6$t3​=9−6+3=6
$t_4 = 16 – 8 + 3 = 11$t4​=16−8+3=11
$t_5 = 25 – 10 + 3 = 18$t5​=25−10+3=18

Answer: $2, 3, 6, 11, 18$2,3,6,11,18

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Q2. Find the 10th and 15th terms of the sequence tn=5n−3t_n = 5n – 3tn​=5n−3 for n≥1n \geq 1n≥1.

Solution:

10th term:
$t_{10} = 5(10) – 3 = 50 – 3 = 47$t10​=5(10)−3=50−3=47

15th term:
$t_{15} = 5(15) – 3 = 75 – 3 = 72$t15​=5(15)−3=75−3=72

Answer:
10th term = 47
15th term = 72

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Q3. Determine whether 97 and 172 are terms of the sequence tn=5n−3t_n = 5n – 3tn​=5n−3 for n≥1n \geq 1n≥1.

Solution:

Check for 97:
$5n – 3 = 97$5n−3=97
$5n = 100 \Rightarrow n = 20$5n=100⇒n=20

✔ 97 is a term

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Check for 172:
$5n – 3 = 172$5n−3=172
$5n = 175 \Rightarrow n = 35$5n=175⇒n=35

✔ 172 is a term

Answer: Both 97 and 172 are terms.

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Q4. Which term of the sequence tn=5n−3t_n = 5n – 3tn​=5n−3 for n≥1n \geq 1n≥1 is 607?

Solution:

$5n – 3 = 607$5n−3=607
$5n = 610 \Rightarrow n = 122$5n=610⇒n=122

Answer: 607 is the 122nd term

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Q5. A sequence is given by the recursive rule t1=−5t_1 = -5t1​=−5, tn+1=tn+3t_{n+1} = t_n + 3tn+1​=tn​+3 for n≥1n \geq 1n≥1. Find the first five terms. Is 52 a term of this sequence? If so, which term is it?

Solution:

First five terms:

$t_1 = -5$t1​=−5
$t_2 = -5 + 3 = -2$t2​=−5+3=−2
$t_3 = -2 + 3 = 1$t3​=−2+3=1
$t_4 = 1 + 3 = 4$t4​=1+3=4
$t_5 = 4 + 3 = 7$t5​=4+3=7

Sequence:
$-5, -2, 1, 4, 7$−5,−2,1,4,7

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Check for 52:

This is an AP with
First term $a = -5$a=−5, common difference $d = 3$d=3

Formula:
$t_n = a + (n-1)d$tn​=a+(n−1)d

$52 = -5 + (n-1)3$52=−5+(n−1)3
$52 + 5 = 3(n-1)$52+5=3(n−1)
$57 = 3(n-1)$57=3(n−1)
$n – 1 = 19 \Rightarrow n = 20$n−1=19⇒n=20

Answer:
Yes, 52 is the 20th term

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Q6. Let T1=1,T2=2,T3=4T_1 = 1, T_2 = 2, T_3 = 4T1​=1,T2​=2,T3​=4, and Tn=Tn−1+Tn−2+Tn−3T_n = T_{n-1} + T_{n-2} + T_{n-3}Tn​=Tn−1​+Tn−2​+Tn−3​ for n≥4n \geq 4n≥4. Find T4,T5,T6,T7,T8T_4, T_5, T_6, T_7, T_8T4​,T5​,T6​,T7​,T8​.

Solution:

$T_4 = 4 + 2 + 1 = 7$T4​=4+2+1=7
$T_5 = 7 + 4 + 2 = 13$T5​=7+4+2=13
$T_6 = 13 + 7 + 4 = 24$T6​=13+7+4=24
$T_7 = 24 + 13 + 7 = 44$T7​=24+13+7=44
$T_8 = 44 + 24 + 13 = 81$T8​=44+24+13=81

Answer:
$T_4 = 7, T_5 = 13, T_6 = 24, T_7 = 44, T_8 = 81$T4​=7,T5​=13,T6​=24,T7​=44,T8​=81

Exercise 8.2

Q1. Find the 10th and 26th terms of the AP: 3, 8, 13, 18, …

Solution:

Given AP: 3, 8, 13, 18, …
First term $a = 3$a=3
Common difference $d = 8 – 3 = 5$d=8−3=5

Formula:
$t_n = a + (n-1)d$tn​=a+(n−1)d

10th term:
$t_{10} = 3 + (10-1)\times5$t10​=3+(10−1)×5
$= 3 + 45 = 48$=3+45=48

26th term:
$t_{26} = 3 + (26-1)\times5$t26​=3+(26−1)×5
$= 3 + 125 = 128$=3+125=128

Answer:
10th term = 48
26th term = 128

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Q2. Which term of the AP: 21, 18, 15, … is –81? Also, is 0 a term of this AP? Give reasons.

Solution:

Given AP: 21, 18, 15, …
$a = 21$a=21, $d = -3$d=−3

Formula:
$t_n = a + (n-1)d$tn​=a+(n−1)d

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For –81:
$-81 = 21 + (n-1)(-3)$−81=21+(n−1)(−3)
$-81 = 21 – 3(n-1)$−81=21−3(n−1)
$-102 = -3(n-1)$−102=−3(n−1)
$n-1 = 34 \Rightarrow n = 35$n−1=34⇒n=35

✔ –81 is the 35th term

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For 0:
$0 = 21 – 3(n-1)$0=21−3(n−1)
$-21 = -3(n-1)$−21=−3(n−1)
$n-1 = 7 \Rightarrow n = 8$n−1=7⇒n=8

✔ 0 is the 8th term

Answer:
–81 is 35th term
0 is 8th term

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Q3. Find the nth term of the AP: 11, 8, 5, 2 … Write the recursive rule for this AP.

Solution:

Given AP: 11, 8, 5, 2 …
$a = 11$a=11, $d = -3$d=−3

nth term:
$t_n = 11 + (n-1)(-3)$tn​=11+(n−1)(−3)
$= 11 – 3(n-1)$=11−3(n−1)

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Recursive rule:
$t_1 = 11$t1​=11
$t_{n+1} = t_n – 3$tn+1​=tn​−3

Answer:
$t_n = 11 – 3(n-1)$tn​=11−3(n−1)
Recursive: $t_{n+1} = t_n – 3$tn+1​=tn​−3

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Q4. An AP consists of 50 terms in which the 3rd term is 12 and the last term is 106. Find the 29th term.

Solution:

Given:
$t_3 = a + 2d = 12$t3​=a+2d=12
$t_{50} = a + 49d = 106$t50​=a+49d=106

Subtract:
$(a + 49d) – (a + 2d) = 106 – 12$(a+49d)−(a+2d)=106−12
$47d = 94 \Rightarrow d = 2$47d=94⇒d=2

Put in first equation:
$a + 2(2) = 12 \Rightarrow a = 8$a+2(2)=12⇒a=8

Now find 29th term:
$t_{29} = 8 + (29-1)\times2$t29​=8+(29−1)×2
$= 8 + 56 = 64$=8+56=64

Answer: 64

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Q5. How many 2-digit numbers are divisible by 3? What is their sum?

Solution:

Smallest 2-digit multiple = 12
Largest = 99

AP: 12, 15, 18, …, 99
$a = 12$a=12, $d = 3$d=3

Number of terms:
$99 = 12 + (n-1)3$99=12+(n−1)3
$87 = 3(n-1) \Rightarrow n-1 = 29 \Rightarrow n = 30$87=3(n−1)⇒n−1=29⇒n=30

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Sum:
$S = \frac{n}{2}(a + l)$S=2n​(a+l)

$= \frac{30}{2}(12 + 99)$=230​(12+99)
$= 15 \times 111 = 1665$=15×111=1665

Answer:
Total numbers = 30
Sum = 1665

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Q6. Harish 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?

Solution:

$a = 500000$a=500000, $d = 20000$d=20000

$700000 = 500000 + (n-1)\times20000$700000=500000+(n−1)×20000
$200000 = (n-1)\times20000$200000=(n−1)×20000
$n-1 = 10 \Rightarrow n = 11$n−1=10⇒n=11

Years after joining = $n-1 = 10$n−1=10

Answer: 10 years

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Q7. A child arranges marbles in rows: 1, 2, 3, … up to 25 rows. How many marbles are used in total?

Solution:

AP: 1, 2, 3, …, 25
$n = 25$n=25, $a = 1$a=1, $l = 25$l=25

Sum:
$S = \frac{n}{2}(a + l)$S=2n​(a+l)

$= \frac{25}{2}(1 + 25)$=225​(1+25)
$= \frac{25}{2} \times 26 = 25 \times 13 = 325$=225​×26=25×13=325

Answer: 325 marbles

Exercise 8.3

Q1. Find the 12th term of a GP with common ratio 2, whose 8th term is 192.

Solution:

Given:
Common ratio $r = 2$r=2
8th term $t_8 = 192$t8​=192

Formula:
$t_n = a \cdot r^{n-1}$tn​=a⋅rn−1

So,
$t_8 = a \cdot 2^7 = 192$t8​=a⋅27=192
$a \cdot 128 = 192$a⋅128=192
$a = \frac{192}{128} = \frac{3}{2}$a=128192​=23​

Now find 12th term:
$t_{12} = a \cdot 2^{11}$t12​=a⋅211
$= \frac{3}{2} \cdot 2048 = 3072$=23​⋅2048=3072

Answer: 3072

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Q2. Find the 10th and nth terms of the GP: 5, 25, 125, …

Solution:

Given GP:
$a = 5$a=5, $r = 5$r=5

Formula:
$t_n = a \cdot r^{n-1}$tn​=a⋅rn−1

10th term:
$t_{10} = 5 \cdot 5^9 = 5^{10} = 9765625$t10​=5⋅59=510=9765625

nth term:
$t_n = 5 \cdot 5^{n-1} = 5^n$tn​=5⋅5n−1=5n

Answer:
10th term = 9765625
nth term = 5n5^n5n

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Q3. A sequence is given by the recursive rule t1=2t_1 = 2t1​=2, tn+1=3tn−2t_{n+1} = 3t_n – 2tn+1​=3tn​−2 for n≥1n \geq 1n≥1. Which term of the sequence is 730?

Solution:

Given:
$t_1 = 2$t1​=2, $t_{n+1} = 3t_n – 2$tn+1​=3tn​−2

Let $t_n – 1 = y_n$tn​−1=yn​

Then:
$y_{n+1} = 3y_n$yn+1​=3yn​

So, this is a GP:
$y_n = 3^{n-1}$yn​=3n−1

Thus,
$t_n = 3^{n-1} + 1$tn​=3n−1+1

Now,
$730 = 3^{n-1} + 1$730=3n−1+1
$729 = 3^{n-1}$729=3n−1
$3^6 = 729$36=729

So,
$n-1 = 6 \Rightarrow n = 7$n−1=6⇒n=7

Answer: 7th term

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Q4. Which term of the GP: 2, 6, 18, … is 4374? Write the explicit formula as well as the recursive formula for the nth term.

Solution:

Given:
$a = 2$a=2, $r = 3$r=3

Formula:
$t_n = 2 \cdot 3^{n-1}$tn​=2⋅3n−1

Now,
$4374 = 2 \cdot 3^{n-1}$4374=2⋅3n−1
$2187 = 3^{n-1}$2187=3n−1
$3^7 = 2187$37=2187

So,
$n-1 = 7 \Rightarrow n = 8$n−1=7⇒n=8

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Explicit formula:
$t_n = 2 \cdot 3^{n-1}$tn​=2⋅3n−1

Recursive formula:
$t_1 = 2$t1​=2,
$t_{n+1} = 3t_n$tn+1​=3tn​

Answer:
4374 is the 8th term

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Q5. A 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.

(i) What height does the ball reach after the 5th bounce?
(ii) What is the total vertical distance travelled by the time it hits the ground for the 6th time?

Solution:

Given:
Initial height = 80 m
Ratio $r = 0.6$r=0.6

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(i) Height after 5th bounce:

$= 80 \times (0.6)^5$=80×(0.6)5
$= 80 \times 0.07776 = 6.2208 \, m$=80×0.07776=6.2208m

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(ii) Total distance till 6th hit:

Distance = First fall + twice sum of first 5 bounces

$= 80 + 2(80(0.6) + 80(0.6)^2 + … + 80(0.6)^5)$=80+2(80(0.6)+80(0.6)2+…+80(0.6)5)

This is GP:
First term = 48, ratio = 0.6, terms = 5

Sum =
$S = 48 \cdot \frac{1 – (0.6)^5}{1 – 0.6}$S=48⋅1−0.61−(0.6)5​
$= 48 \cdot \frac{1 – 0.07776}{0.4}$=48⋅0.41−0.07776​
$= 48 \cdot 2.3056 = 110.6688$=48⋅2.3056=110.6688

Total distance:
$= 80 + 2(110.6688) = 301.3376 \, m$=80+2(110.6688)=301.3376m

Answer:
(i) 6.2208 m
(ii) 301.34 m (approx.)

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Q6. Which term of the sequence 2, 2√2, 4, … is 128?

Solution:

This is GP:
$a = 2$a=2, $r = \sqrt{2}$r=2​

Formula:
$t_n = 2(\sqrt{2})^{n-1}$tn​=2(2​)n−1

$128 = 2(\sqrt{2})^{n-1}$128=2(2​)n−1
$64 = (\sqrt{2})^{n-1}$64=(2​)n−1

$(\sqrt{2})^{n-1} = (2^{1/2})^{n-1} = 2^{(n-1)/2}$(2​)n−1=(21/2)n−1=2(n−1)/2

So,
$2^{(n-1)/2} = 2^6$2(n−1)/2=26

$\frac{n-1}{2} = 6 \Rightarrow n-1 = 12 \Rightarrow n = 13$2n−1​=6⇒n−1=12⇒n=13

Answer: 13th term

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Q7. Sierpiński square carpet problem (pattern-based)

(i) Number of red squares:
Stage 0 = 1
Stage 1 = 8
Stage 2 = 64
Stage 3 = 512

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(ii) Stage 4 = $8^4 = 4096$84=4096
Stage 5 = $8^5 = 32768$85=32768

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(iii) Rule:
Explicit: $T_n = 8^n$Tn​=8n
Recursive: $T_1 = 8$T1​=8, $T_{n+1} = 8T_n$Tn+1​=8Tn​

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(iv) Area pattern:
Stage 1 = $\frac{8}{9}$98​
Stage 2 = $\left(\frac{8}{9}\right)^2$(98​)2
Stage 3 = $\left(\frac{8}{9}\right)^3$(98​)3

General:
Area = $\left(\frac{8}{9}\right)^n$(98​)n

As $n \to \infty$n→∞, area → 0

End-of-Chapter Exercises

Q1. Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.

Solution:

Given:
11th term → $t_{11} = a + 10d = 38$t11​=a+10d=38
16th term → $t_{16} = a + 15d = 73$t16​=a+15d=73

Subtracting:
$(a + 15d) – (a + 10d) = 73 – 38$(a+15d)−(a+10d)=73−38
$5d = 35 \Rightarrow d = 7$5d=35⇒d=7

Substitute in first equation:
$a + 10(7) = 38 \Rightarrow a = -32$a+10(7)=38⇒a=−32

Now,
$t_{31} = a + 30d = -32 + 210 = 178$t31​=a+30d=−32+210=178

Answer: 178

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Q2. Determine the AP whose third term is 16 and whose 7th term exceeds the 5th term by 12.

Solution:

Given:
Third term → $t_3 = a + 2d = 16$t3​=a+2d=16

Also,
$t_7 – t_5 = 12$t7​−t5​=12

So,
$(a + 6d) – (a + 4d) = 12$(a+6d)−(a+4d)=12
$2d = 12 \Rightarrow d = 6$2d=12⇒d=6

Now,
$a + 2(6) = 16 \Rightarrow a = 4$a+2(6)=16⇒a=4

AP: $4, 10, 16, 22, …$4,10,16,22,…

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Q3. How 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.)

Solution:

Smallest 3-digit multiple of 7 = 105
Largest = 994

AP: $105, 112, 119, …, 994$105,112,119,…,994

$a = 105, d = 7$a=105,d=7

$994 = 105 + (n-1)7$994=105+(n−1)7
$889 = 7(n-1) \Rightarrow n-1 = 127 \Rightarrow n = 128$889=7(n−1)⇒n−1=127⇒n=128

Answer: 128 numbers

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Q4. How 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.)

Solution:

Smallest multiple = 12
Largest = 248

AP: $12, 16, 20, …, 248$12,16,20,…,248

$a = 12, d = 4$a=12,d=4

$248 = 12 + (n-1)4$248=12+(n−1)4
$236 = 4(n-1) \Rightarrow n = 60$236=4(n−1)⇒n=60

Answer: 60 multiples

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Q5. Find a GP for which the sum of the first two terms is – 4 and the fifth term is 4 times the third term.

Solution:

Let first term = $a$a, ratio = $r$r

Condition 1:
$a + ar = -4 \Rightarrow a(1+r) = -4$a+ar=−4⇒a(1+r)=−4

Condition 2:
$ar^4 = 4(ar^2) \Rightarrow r^2 = 4 \Rightarrow r = 2 \text{ or } -2$ar4=4(ar2)⇒r2=4⇒r=2 or −2

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Case 1: $r = 2$r=2
$a(3) = -4 \Rightarrow a = -\frac{4}{3}$a(3)=−4⇒a=−34​

GP: $-\frac{4}{3}, -\frac{8}{3}, …$−34​,−38​,…

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Case 2: $r = -2$r=−2
$a(-1) = -4 \Rightarrow a = 4$a(−1)=−4⇒a=4

GP: $4, -8, 16, …$4,−8,16,…

Answer: Two possible GPs

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Q6. Find all possible ways of expressing 100 as the sum of consecutive natural numbers.

Solution:

Try sequences:

$100 = 18 + 19 + 20 + 21 + 22$100=18+19+20+21+22

$100 = 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16$100=9+10+11+12+13+14+15+16

Answer: 2 ways

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Q7. The 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?

Solution:

This is a GP:
$a = 30, r = 2$a=30,r=2

After 2 hours:
$30 \times 2^2 = 120$30×22=120

After 4 hours:
$30 \times 2^4 = 480$30×24=480

After n hours:
$30 \times 2^n$30×2n

Answer:
120, 480, $30 \times 2^n$30×2n

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Q8. The 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.

Solution:

$t_4 + t_8 = (a+3d)+(a+7d) = 2a+10d = 24$t4​+t8​=(a+3d)+(a+7d)=2a+10d=24

$t_6 + t_{10} = (a+5d)+(a+9d) = 2a+14d = 44$t6​+t10​=(a+5d)+(a+9d)=2a+14d=44

Subtract:
$4d = 20 \Rightarrow d = 5$4d=20⇒d=5

$2a + 50 = 24 \Rightarrow a = -13$2a+50=24⇒a=−13

First three terms:
$-13, -8, -3$−13,−8,−3

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Q9. Find the smallest value of n such that the sum of the first n natural numbers is greater than 1,000.

Solution:

$S = \frac{n(n+1)}{2} > 1000$S=2n(n+1)​>1000

$n(n+1) > 2000$n(n+1)>2000

Try:
$n = 44 → 990$n=44→990
$n = 45 → 1035$n=45→1035

Answer: 45

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Q10. Which term of the GP: 2, 8, 32, … is 131072? Write the explicit formula as well as the recursive formula for the nth term.

Solution:

$a = 2, r = 4$a=2,r=4

$131072 = 2 \cdot 4^{n-1}$131072=2⋅4n−1
$65536 = 4^{n-1} = 2^{16}$65536=4n−1=216

$2(n-1) = 16 \Rightarrow n = 9$2(n−1)=16⇒n=9

Explicit: $t_n = 2 \cdot 4^{n-1}$tn​=2⋅4n−1
Recursive: $t_{n+1} = 4t_n$tn+1​=4tn​

Answer: 9th term

13\frac{13}{ }13​ and their product is –1. Find the common ratio and the terms.

(Note: The question seems slightly misprinted; taking standard form where sum = 13.)

Solution:

Let the three terms be:
$\frac{a}{r},\ a,\ ar$ra​, a, ar

Product =
$\frac{a}{r} \cdot a \cdot ar = a^3$ra​⋅a⋅ar=a3

Given:
$a^3 = -1 \Rightarrow a = -1$a3=−1⇒a=−1

Now sum:
$\frac{-1}{r} -1 – r = 13$r−1​−1−r=13

Multiply by $r$r:
$-1 – r – r^2 = 13r$−1−r−r2=13r

$r^2 + 14r + 1 = 0$r2+14r+1=0

Solve:
$r = -7 \pm 4\sqrt{3}$r=−7±43​

Now terms:
$\frac{-1}{r},\ -1,\ -r$r−1​, −1, −r

Answer:
Common ratio $r = -7 \pm 4\sqrt{3}$r=−7±43​
Terms accordingly

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Q12. If the 4th, 10th and 16th terms of a GP are x, y and z respectively, prove that x, y, z are in GP.

Solution:

Let first term = $a$a, ratio = $r$r

Then:
$x = ar^3$x=ar3
$y = ar^9$y=ar9
$z = ar^{15}$z=ar15

Now check:
$y^2 = (ar^9)^2 = a^2 r^{18}$y2=(ar9)2=a2r18

$xz = (ar^3)(ar^{15}) = a^2 r^{18}$xz=(ar3)(ar15)=a2r18

So,
$y^2 = xz$y2=xz

Hence, $x, y, z$x,y,z are in GP

Answer: Proved

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Q13. The 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.

Solution:

Let terms be:
$\frac{a}{r},\ a,\ ar$ra​, a, ar

Sum:
$\frac{a}{r} + a + ar = 26$ra​+a+ar=26

Squares:
$\frac{a^2}{r^2} + a^2 + a^2 r^2 = 364$r2a2​+a2+a2r2=364

Try simple GP:

Take $a = 6, r = 3$a=6,r=3

Terms: $2, 6, 18$2,6,18

Check:
Sum = $2+6+18 = 26$2+6+18=26 ✔
Squares = $4+36+324 = 364$4+36+324=364 ✔

Answer: 2, 6, 18

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Q14. Suppose P1=1,P2=2P_1 = 1, P_2 = 2P1​=1,P2​=2 and for n>2n > 2n>2, Pn=P1+P2+…+Pn−1+1P_n = P_1 + P_2 + … + P_{n-1} + 1Pn​=P1​+P2​+…+Pn−1​+1. Find the values of P1,P2,…,P8P_1, P_2, …, P_8P1​,P2​,…,P8​. Can you find a simpler recursive formula for PnP_nPn​? Can you give an explicit formula?

Solution:

Given:
$P_1 = 1, P_2 = 2$P1​=1,P2​=2

Now:
$P_3 = 1+2+1 = 4$P3​=1+2+1=4
$P_4 = 1+2+4+1 = 8$P4​=1+2+4+1=8
$P_5 = 1+2+4+8+1 = 16$P5​=1+2+4+8+1=16
$P_6 = 32$P6​=32
$P_7 = 64$P7​=64
$P_8 = 128$P8​=128

So sequence:
$1, 2, 4, 8, 16, 32, 64, 128$1,2,4,8,16,32,64,128

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Simpler recursive rule:
$P_n = 2P_{n-1}$Pn​=2Pn−1​

Explicit formula:
$P_n = 2^{n-1}$Pn​=2n−1

Answer:
Values up to $P_8 = 128$P8​=128,
Recursive: $P_n = 2P_{n-1}$Pn​=2Pn−1​,
Explicit: $2^{n-1}$2n−1

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Q15. Suppose W1=1,W2=2W_1 = 1, W_2 = 2W1​=1,W2​=2 and for n>2n > 2n>2, Wn=W1+W2+…+Wn−2+2W_n = W_1 + W_2 + … + W_{n-2} + 2Wn​=W1​+W2​+…+Wn−2​+2. Find the values of W1,W2,…,W8W_1, W_2 , …, W_8W1​,W2​,…,W8​. Do you recognise this sequence?

Solution:

Given:
$W_1 = 1, W_2 = 2$W1​=1,W2​=2

Now:
$W_3 = 1 + 2 = 3$W3​=1+2=3
$W_4 = 1 + 2 + 2 = 5$W4​=1+2+2=5
$W_5 = 1 + 2 + 3 + 2 = 8$W5​=1+2+3+2=8
$W_6 = 13$W6​=13
$W_7 = 21$W7​=21
$W_8 = 34$W8​=34

Sequence:
$1, 2, 3, 5, 8, 13, 21, 34$1,2,3,5,8,13,21,34

This is Fibonacci sequence

Answer:
Sequence is Fibonacci pattern