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Sequences and Series

CBSE · Class 11 · Applied Mathematics

NCERT Solutions for Sequences and Series — CBSE Class 11 Applied Mathematics.

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Check Your Progress 1 — Multiple Choice Questions

1If for a sequence Sn=2n−3S_n = 2n - 3, then the common difference is:
(a) −1-1 (b) 2 (c) −2-2 (d) 3Show solution
Given: Sn=2n−3S_n = 2n - 3

Concept: The nthn^{\text{th}} term of a sequence is an=Sn−Sn−1a_n = S_n - S_{n-1}.

Working:
an=Sn−Sn−1=(2n−3)−(2(n−1)−3)=(2n−3)−(2n−5)=2a_n = S_n - S_{n-1} = (2n-3) - (2(n-1)-3) = (2n-3)-(2n-5) = 2

Since every term equals 2 (a constant), the common difference d=an−an−1=2d = a_n - a_{n-1} = 2.

Answer: (b) 2

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2The number of integers from 100 to 500 that are divisible by 5 are:
(a) 80 (b) 81 (c) 75 (d) none of theseShow solution
Given: Integers from 100 to 500 divisible by 5.

Concept: These form an A.P.: 100,105,110,…,500100, 105, 110, \ldots, 500 with a=100a = 100, d=5d = 5, l=500l = 500.

Working:
l=a+(n−1)dl = a + (n-1)d
500=100+(n−1)×5500 = 100 + (n-1)\times 5
400=5(n−1)400 = 5(n-1)
n−1=80⇒n=81n-1 = 80 \Rightarrow n = 81

Answer: (b) 81

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3The number of two digit numbers divisible by 6 are:
(a) 24 (b) 14 (c) 15 (d) 20Show solution
Given: Two-digit numbers divisible by 6.

Concept: These form an A.P.: 12,18,24,…,9612, 18, 24, \ldots, 96 with a=12a = 12, d=6d = 6, l=96l = 96.

Working:
l=a+(n−1)dl = a + (n-1)d
96=12+(n−1)×696 = 12 + (n-1)\times 6
84=6(n−1)84 = 6(n-1)
n−1=14⇒n=15n-1 = 14 \Rightarrow n = 15

Answer: (c) 15

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4If the fourth term of an A.P. is 4, then the sum of its 7 terms is:
(a) 28 (b) 26 (c) 32 (d) none of theseShow solution
Given: a4=4a_4 = 4

Concept: For an A.P., Sn=n2(a+l)S_n = \dfrac{n}{2}(a + l). Also, the middle term of a 7-term A.P. is a4a_4, and S7=7×a4S_7 = 7 \times a_4.

Working:
S7=72[2a+6d]=7(a+3d)=7×a4=7×4=28S_7 = \frac{7}{2}[2a + 6d] = 7(a + 3d) = 7 \times a_4 = 7 \times 4 = 28

Answer: (a) 28

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5Which of the following terms are NOT a term of the A.P. −3,−7,−11,−15,…,−403,…,−799-3, -7, -11, -15, \ldots, -403, \ldots, -799?
(a) −500-500 (b) −399-399 (c) −503-503 (d) −51-51Show solution
Given: A.P. with a=−3a = -3, d=−4d = -4.

Concept: an=a+(n−1)d=−3+(n−1)(−4)=−3−4n+4=1−4na_n = a + (n-1)d = -3 + (n-1)(-4) = -3 - 4n + 4 = 1 - 4n.

For a number kk to be a term: k=1−4n⇒n=1−k4k = 1 - 4n \Rightarrow n = \dfrac{1-k}{4} must be a positive integer.

Checking each option:
- (a) k=−500k = -500: n=1−(−500)4=5014=125.25n = \dfrac{1-(-500)}{4} = \dfrac{501}{4} = 125.25 — not an integer, so −500-500 is NOT a term.
- (b) k=−399k = -399: n=4004=100n = \dfrac{400}{4} = 100 — integer, so it IS a term.
- (c) k=−503k = -503: n=5044=126n = \dfrac{504}{4} = 126 — integer, so it IS a term.
- (d) k=−51k = -51: n=524=13n = \dfrac{52}{4} = 13 — integer, so it IS a term.

Answer: **(a) −500-500**

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Check Your Progress 1 — Short Answer Type Questions

6If each term in a given A.P. is doubled, then is the new sequence obtained an A.P.? If yes, then find its common difference.Show solution
Given: An A.P. with first term aa and common difference dd: a, a+d, a+2d, …a,\ a+d,\ a+2d,\ \ldots

New sequence (each term doubled): 2a, 2(a+d), 2(a+2d), …2a,\ 2(a+d),\ 2(a+2d),\ \ldots

Check for A.P.:
Difference between consecutive terms:
2(a+d)−2a=2d2(a+d) - 2a = 2d
2(a+2d)−2(a+d)=2d2(a+2d) - 2(a+d) = 2d

The difference is constant =2d= 2d.

Conclusion: Yes, the new sequence is an A.P. with common difference **2d2d** (twice the common difference of the original A.P.).

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7Find the middle term in the A.P. 20,16,12,…,−17620, 16, 12, \ldots, -176.Show solution
Given: A.P. with a=20a = 20, d=−4d = -4, l=−176l = -176.

Step 1: Find number of terms.
l=a+(n−1)dl = a + (n-1)d
−176=20+(n−1)(−4)-176 = 20 + (n-1)(-4)
−196=−4(n−1)-196 = -4(n-1)
n−1=49⇒n=50n-1 = 49 \Rightarrow n = 50

Step 2: Find middle terms.
Since n=50n = 50 (even), there are two middle terms: a25a_{25} and a26a_{26}.

a25=20+24(−4)=20−96=−76a_{25} = 20 + 24(-4) = 20 - 96 = -76
a26=20+25(−4)=20−100=−80a_{26} = 20 + 25(-4) = 20 - 100 = -80

Answer: The middle terms are a25=−76a_{25} = -76 and a26=−80a_{26} = -80.

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8In an A.P., if a4:a7=2:3a_4 : a_7 = 2 : 3, then find a5:a8a_5 : a_8.Show solution
Given: a4:a7=2:3a_4 : a_7 = 2 : 3

Let first term =a= a, common difference =d= d.

a+3da+6d=23\frac{a + 3d}{a + 6d} = \frac{2}{3}
3(a+3d)=2(a+6d)3(a + 3d) = 2(a + 6d)
3a+9d=2a+12d3a + 9d = 2a + 12d
a=3da = 3d

**Now find a5:a8a_5 : a_8:**
a5=a+4d=3d+4d=7da_5 = a + 4d = 3d + 4d = 7d
a8=a+7d=3d+7d=10da_8 = a + 7d = 3d + 7d = 10d
a5:a8=7d:10d=7:10a_5 : a_8 = 7d : 10d = \boxed{7 : 10}

*(Note: The answer key states a6:a8=4:5a_6:a_8 = 4:5; the question asks for a5:a8=7:10a_5:a_8 = 7:10.)*

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9Find the sum of integers from 100 to 500 that are divisible by 2 and 3.Show solution
Given: Integers from 100 to 500 divisible by both 2 and 3, i.e., divisible by LCM(2,3)=6(2,3) = 6.

A.P.: 102,108,114,…,498102, 108, 114, \ldots, 498 with a=102a = 102, d=6d = 6, l=498l = 498.

**Step 1: Find nn.**
498=102+(n−1)×6498 = 102 + (n-1)\times 6
396=6(n−1)396 = 6(n-1)
n=67n = 67

Step 2: Find sum.
S=n2(a+l)=672(102+498)=672×600=67×300=20100S = \frac{n}{2}(a + l) = \frac{67}{2}(102 + 498) = \frac{67}{2} \times 600 = 67 \times 300 = \boxed{20100}

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10Find the sum of 20 terms of the A.P. whose nthn^{\text{th}} term is 2n+12n + 1.Show solution
Given: an=2n+1a_n = 2n + 1

First term: a1=2(1)+1=3a_1 = 2(1)+1 = 3

Common difference: d=a2−a1=5−3=2d = a_2 - a_1 = 5 - 3 = 2

Sum of 20 terms:
S20=202[2a+19d]=10[2(3)+19(2)]=10[6+38]=10×44=440S_{20} = \frac{20}{2}[2a + 19d] = 10[2(3) + 19(2)] = 10[6 + 38] = 10 \times 44 = \boxed{440}

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Check Your Progress 1 — Long Answer Type Questions

11Insert nn arithmetic means between 1 and 31 such that the ratio of the 7th7^{\text{th}} mean and the (n−1)th(n-1)^{\text{th}} mean is 5:95:9. Find the value of nn and the resulting A.P.Show solution
Given: nn A.M.s are inserted between 1 and 31. So the full sequence is:
1,A1,A2,…,An,311, A_1, A_2, \ldots, A_n, 31
Total terms =n+2= n + 2, first term a=1a = 1, last term =31= 31.

Common difference:
d=31−1n+1=30n+1d = \frac{31 - 1}{n + 1} = \frac{30}{n+1}

**The kthk^{\text{th}} arithmetic mean:** Ak=1+kdA_k = 1 + kd

A7=1+7d,An−1=1+(n−1)dA_7 = 1 + 7d, \quad A_{n-1} = 1 + (n-1)d

Given ratio:
A7An−1=59\frac{A_7}{A_{n-1}} = \frac{5}{9}
1+7d1+(n−1)d=59\frac{1 + 7d}{1 + (n-1)d} = \frac{5}{9}

Substitute d=30n+1d = \dfrac{30}{n+1}:
1+210n+11+30(n−1)n+1=59\frac{1 + \dfrac{210}{n+1}}{1 + \dfrac{30(n-1)}{n+1}} = \frac{5}{9}

Multiply numerator and denominator by (n+1)(n+1):
(n+1)+210(n+1)+30(n−1)=59\frac{(n+1) + 210}{(n+1) + 30(n-1)} = \frac{5}{9}
n+211n+1+30n−30=59\frac{n + 211}{n + 1 + 30n - 30} = \frac{5}{9}
n+21131n−29=59\frac{n + 211}{31n - 29} = \frac{5}{9}
9(n+211)=5(31n−29)9(n + 211) = 5(31n - 29)
9n+1899=155n−1459n + 1899 = 155n - 145
2044=146n2044 = 146n
n=14n = 14

Common difference: d=3015=2d = \dfrac{30}{15} = 2

Resulting A.P.: 1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,311, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31

Answer: n=14n = 14; the A.P. is 1,3,5,7,…,311, 3, 5, 7, \ldots, 31 with d=2d = 2.

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12Find the sum of the following series:
(a) 4+7+10+…4 + 7 + 10 + \dots to 100 terms
(b) 1+43+53+2+…1 + \dfrac{4}{3} + \dfrac{5}{3} + 2 + \dots to 19 terms
(c) 0.5+0.51+0.52+…0.5 + 0.51 + 0.52 + \dots to 1000 termsShow solution
(a) 4+7+10+…4 + 7 + 10 + \dots to 100 terms

a=4a = 4, d=3d = 3, n=100n = 100
S100=1002[2(4)+99(3)]=50[8+297]=50×305=15250S_{100} = \frac{100}{2}[2(4) + 99(3)] = 50[8 + 297] = 50 \times 305 = \boxed{15250}

(b) 1+43+53+2+…1 + \dfrac{4}{3} + \dfrac{5}{3} + 2 + \dots to 19 terms

a=1a = 1, d=43−1=13d = \dfrac{4}{3} - 1 = \dfrac{1}{3}, n=19n = 19
S19=192[2(1)+18(13)]=192[2+6]=192×8=76S_{19} = \frac{19}{2}\left[2(1) + 18\left(\frac{1}{3}\right)\right] = \frac{19}{2}\left[2 + 6\right] = \frac{19}{2} \times 8 = \boxed{76}

(c) 0.5+0.51+0.52+…0.5 + 0.51 + 0.52 + \dots to 1000 terms

a=0.5a = 0.5, d=0.01d = 0.01, n=1000n = 1000
S1000=10002[2(0.5)+999(0.01)]=500[1+9.99]=500×10.99=5495S_{1000} = \frac{1000}{2}[2(0.5) + 999(0.01)] = 500[1 + 9.99] = 500 \times 10.99 = \boxed{5495}

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13If the sum of terms of an A.P. is 72, find the number of terms given that the first term of the sequence is 17 and common difference is −2-2.Show solution
Given: Sn=72S_n = 72, a=17a = 17, d=−2d = -2

Formula:
Sn=n2[2a+(n−1)d]S_n = \frac{n}{2}[2a + (n-1)d]
72=n2[2(17)+(n−1)(−2)]72 = \frac{n}{2}[2(17) + (n-1)(-2)]
144=n[34−2n+2]144 = n[34 - 2n + 2]
144=n[36−2n]144 = n[36 - 2n]
144=36n−2n2144 = 36n - 2n^2
2n2−36n+144=02n^2 - 36n + 144 = 0
n2−18n+72=0n^2 - 18n + 72 = 0
(n−6)(n−12)=0(n - 6)(n - 12) = 0
n=6orn=12n = 6 \quad \text{or} \quad n = 12

Verification:
- S6=62[34+5(−2)]=3×24=72S_6 = \frac{6}{2}[34 + 5(-2)] = 3 \times 24 = 72 ✓
- S12=122[34+11(−2)]=6×12=72S_{12} = \frac{12}{2}[34 + 11(-2)] = 6 \times 12 = 72 ✓

Answer: n=6n = 6 or n=12n = 12.

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14If the sum of first pp terms of an A.P. is equal to the sum of the first qq terms, then prove that the sum of the first (p+q)(p + q) terms is zero.Show solution
Given: Sp=SqS_p = S_q, where Sn=n2[2a+(n−1)d]S_n = \dfrac{n}{2}[2a + (n-1)d].

To prove: Sp+q=0S_{p+q} = 0

Proof:
Sp=SqS_p = S_q
p2[2a+(p−1)d]=q2[2a+(q−1)d]\frac{p}{2}[2a + (p-1)d] = \frac{q}{2}[2a + (q-1)d]
p[2a+(p−1)d]=q[2a+(q−1)d]p[2a + (p-1)d] = q[2a + (q-1)d]
2ap+p(p−1)d=2aq+q(q−1)d2ap + p(p-1)d = 2aq + q(q-1)d
2a(p−q)+d[p(p−1)−q(q−1)]=02a(p - q) + d[p(p-1) - q(q-1)] = 0
2a(p−q)+d[p2−p−q2+q]=02a(p - q) + d[p^2 - p - q^2 + q] = 0
2a(p−q)+d[(p2−q2)−(p−q)]=02a(p - q) + d[(p^2 - q^2) - (p - q)] = 0
2a(p−q)+d(p−q)(p+q−1)=02a(p - q) + d(p - q)(p + q - 1) = 0

Since p≠qp \neq q, divide by (p−q)(p - q):
2a+d(p+q−1)=0…(1)2a + d(p + q - 1) = 0 \quad \ldots (1)

**Now compute Sp+qS_{p+q}:**
Sp+q=p+q2[2a+(p+q−1)d]S_{p+q} = \frac{p+q}{2}[2a + (p+q-1)d]

From (1): 2a+(p+q−1)d=02a + (p+q-1)d = 0

∴Sp+q=p+q2×0=0\therefore S_{p+q} = \frac{p+q}{2} \times 0 = 0

Hence proved. ■\blacksquare

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15Find the first negative term in the given A.P.: 19, 915, 875, …19,\ \dfrac{91}{5},\ \dfrac{87}{5},\ \dotsShow solution
Given: A.P.: 19, 915, 875, …19,\ \dfrac{91}{5},\ \dfrac{87}{5},\ \ldots

a=19=955a = 19 = \dfrac{95}{5}, d=915−955=−45d = \dfrac{91}{5} - \dfrac{95}{5} = -\dfrac{4}{5}

General term:
an=a+(n−1)d=955+(n−1)(−45)=95−4(n−1)5=99−4n5a_n = a + (n-1)d = \frac{95}{5} + (n-1)\left(-\frac{4}{5}\right) = \frac{95 - 4(n-1)}{5} = \frac{99 - 4n}{5}

For first negative term: an<0a_n < 0
99−4n5<0\frac{99 - 4n}{5} < 0
99−4n<099 - 4n < 0
n>994=24.75n > \frac{99}{4} = 24.75

So the first negative term is at n=25n = 25.

a25=99−4(25)5=99−1005=−15a_{25} = \frac{99 - 4(25)}{5} = \frac{99 - 100}{5} = \frac{-1}{5}

Answer: The first negative term is the 25th25^{\text{th}} term =−15= -\dfrac{1}{5}.

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16If pthp^{\text{th}}, qthq^{\text{th}} and rthr^{\text{th}} terms of an A.P. are aa, bb, cc respectively, then prove that (a−b)r+(b−c)p+(c−a)q=0(a - b)r + (b - c)p + (c - a)q = 0.Show solution
Given: Let first term of A.P. be AA and common difference be DD.

a=A+(p−1)D,b=A+(q−1)D,c=A+(r−1)Da = A + (p-1)D, \quad b = A + (q-1)D, \quad c = A + (r-1)D

Compute differences:
a−b=(p−1)D−(q−1)D=(p−q)Da - b = (p-1)D - (q-1)D = (p-q)D
b−c=(q−r)Db - c = (q-r)D
c−a=(r−p)Dc - a = (r-p)D

LHS:
(a−b)r+(b−c)p+(c−a)q(a-b)r + (b-c)p + (c-a)q
=D[(p−q)r+(q−r)p+(r−p)q]= D[(p-q)r + (q-r)p + (r-p)q]
=D[pr−qr+pq−pr+qr−pq]= D[pr - qr + pq - pr + qr - pq]
=D[0]= D[0]
=0=RHS= 0 = \text{RHS}

Hence proved. ■\blacksquare

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17Find the middle terms in the A.P. 5,8,11,14,…5, 8, 11, 14, \ldots whose last term is 95.Show solution
Given: a=5a = 5, d=3d = 3, l=95l = 95.

**Step 1: Find nn.**
95=5+(n−1)(3)⇒90=3(n−1)⇒n=3195 = 5 + (n-1)(3) \Rightarrow 90 = 3(n-1) \Rightarrow n = 31

Step 2: Middle term.
Since n=31n = 31 (odd), there is one middle term: a16a_{16}.

a16=5+15×3=5+45=50a_{16} = 5 + 15 \times 3 = 5 + 45 = \boxed{50}

Answer: The middle term is the 16th16^{\text{th}} term =50= 50.

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18In an A.P., if pthp^{\text{th}} term is 1q\dfrac{1}{q} and qthq^{\text{th}} term is 1p\dfrac{1}{p}, then prove that the sum of first pqpq terms is 12(pq+1)\dfrac{1}{2}(pq + 1), where p≠qp \neq q.Show solution
Given: ap=1qa_p = \dfrac{1}{q} and aq=1pa_q = \dfrac{1}{p}.

**Step 1: Find aa and dd.**
a+(p−1)d=1q…(1)a + (p-1)d = \frac{1}{q} \quad \ldots (1)
a+(q−1)d=1p…(2)a + (q-1)d = \frac{1}{p} \quad \ldots (2)

Subtracting (2) from (1):
(p−q)d=1q−1p=p−qpq(p-q)d = \frac{1}{q} - \frac{1}{p} = \frac{p-q}{pq}
d=1pqd = \frac{1}{pq}

From (1): a=1q−(p−1)1pq=p−(p−1)pq=1pqa = \dfrac{1}{q} - (p-1)\dfrac{1}{pq} = \dfrac{p - (p-1)}{pq} = \dfrac{1}{pq}

**Step 2: Sum of pqpq terms.**
Spq=pq2[2a+(pq−1)d]S_{pq} = \frac{pq}{2}[2a + (pq-1)d]
=pq2[2pq+(pq−1)⋅1pq]= \frac{pq}{2}\left[\frac{2}{pq} + (pq-1)\cdot\frac{1}{pq}\right]
=pq2⋅2+pq−1pq= \frac{pq}{2} \cdot \frac{2 + pq - 1}{pq}
=pq+12=12(pq+1)= \frac{pq+1}{2} = \frac{1}{2}(pq+1)

Hence proved. ■\blacksquare

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19The sum of nn terms of two A.P.s are in the ratio 5n+4:9n+65n + 4 : 9n + 6. Find the ratio of their 18th18^{\text{th}} terms.Show solution
Given: SnSn′=5n+49n+6\dfrac{S_n}{S_n'} = \dfrac{5n+4}{9n+6}

Concept: The ratio of kthk^{\text{th}} terms equals the ratio of sums when n=2k−1n = 2k - 1.

For the 18th18^{\text{th}} term, put n=2(18)−1=35n = 2(18) - 1 = 35:

a18a18′=5(35)+49(35)+6=175+4315+6=179321\frac{a_{18}}{a_{18}'} = \frac{5(35)+4}{9(35)+6} = \frac{175+4}{315+6} = \frac{179}{321}

Answer: The ratio of their 18th18^{\text{th}} terms is 179:321\boxed{179 : 321}.

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20On a certain day in a hospital, during covid crisis, the patients in the OPD were 1000. Due to efforts of the doctors and health care warriors and precautions taken by general public, numbers declined by 50 per day. As per the decline in the number of patients, do you think that there would be a day with no patients in the OPD? If yes, which day would it be from the day when there were 1000 patients?Show solution
Given: a=1000a = 1000, d=−50d = -50.

For zero patients: an=0a_n = 0
1000+(n−1)(−50)=01000 + (n-1)(-50) = 0
1000=50(n−1)1000 = 50(n-1)
n−1=20⇒n=21n - 1 = 20 \Rightarrow n = 21

Answer: Yes, on the 21st day from the day when there were 1000 patients, there would be no COVID patients in the OPD.

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Check Your Progress 2

1(i)Find the indicated term in the Geometric Progression: 4,12,36,…4, 12, 36, \ldots, 5th term.Show solution
Given: G.P. with a=4a = 4, r=124=3r = \dfrac{12}{4} = 3, find a5a_5.

a5=ar5−1=4×34=4×81=324a_5 = ar^{5-1} = 4 \times 3^4 = 4 \times 81 = \boxed{324}

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1(ii)Find the 4th term and nth term of the G.P.: 3,−1,13,−19,…3, -1, \dfrac{1}{3}, -\dfrac{1}{9}, \ldotsShow solution
Given: a=3a = 3, r=−13r = \dfrac{-1}{3}.

4th term:
a4=ar3=3×(−13)3=3×(−127)=−19a_4 = ar^3 = 3 \times \left(-\frac{1}{3}\right)^3 = 3 \times \left(-\frac{1}{27}\right) = -\frac{1}{9}

nth term:
an=arn−1=3×(−13)n−1=3×(−1)n−13n−1=(−1)n−13n−2a_n = ar^{n-1} = 3 \times \left(-\frac{1}{3}\right)^{n-1} = 3 \times \frac{(-1)^{n-1}}{3^{n-1}} = \frac{(-1)^{n-1}}{3^{n-2}}

Answer: a4=−19a_4 = -\dfrac{1}{9}; an=(−1)n−13n−2a_n = \dfrac{(-1)^{n-1}}{3^{n-2}}

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2(i)Which term of the sequence 5,10,20,40,…5, 10, 20, 40, \ldots is 5120?Show solution
Given: a=5a = 5, r=2r = 2, an=5120a_n = 5120.

5×2n−1=51205 \times 2^{n-1} = 5120
2n−1=1024=2102^{n-1} = 1024 = 2^{10}
n−1=10⇒n=11n - 1 = 10 \Rightarrow n = 11

Answer: 5120 is the 11th term.

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2(ii)Which term of the sequence 2,22,4,…2, 2\sqrt{2}, 4, \ldots is 128?Show solution
Given: a=2a = 2, r=222=2r = \dfrac{2\sqrt{2}}{2} = \sqrt{2}, an=128a_n = 128.

2×(2)n−1=1282 \times (\sqrt{2})^{n-1} = 128
(2)n−1=64=26=(2)12(\sqrt{2})^{n-1} = 64 = 2^6 = (\sqrt{2})^{12}
n−1=12⇒n=13n - 1 = 12 \Rightarrow n = 13

Answer: 128 is the 13th term.

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2(iii)Which term of the sequence 2,1,12,14,…2, 1, \dfrac{1}{2}, \dfrac{1}{4}, \ldots is 1128\dfrac{1}{128}?Show solution
Given: a=2a = 2, r=12r = \dfrac{1}{2}, an=1128a_n = \dfrac{1}{128}.

2×(12)n−1=11282 \times \left(\frac{1}{2}\right)^{n-1} = \frac{1}{128}
(12)n−1=1256=(12)8\left(\frac{1}{2}\right)^{n-1} = \frac{1}{256} = \left(\frac{1}{2}\right)^8
n−1=8⇒n=9n - 1 = 8 \Rightarrow n = 9

Answer: 1128\dfrac{1}{128} is the 9th term.

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3(i)Find the sum to 6 terms of the G.P.: 3,3,33,…\sqrt{3}, 3, 3\sqrt{3}, \ldotsShow solution
Given: a=3a = \sqrt{3}, r=33=3r = \dfrac{3}{\sqrt{3}} = \sqrt{3}, n=6n = 6.

S6=a(r6−1)r−1=3[(3)6−1]3−1=3(27−1)3−1=2633−1S_6 = \frac{a(r^6 - 1)}{r - 1} = \frac{\sqrt{3}\left[(\sqrt{3})^6 - 1\right]}{\sqrt{3} - 1} = \frac{\sqrt{3}(27 - 1)}{\sqrt{3} - 1} = \frac{26\sqrt{3}}{\sqrt{3} - 1}

Rationalise:
=263(3+1)(3−1)(3+1)=263(3+1)2=133(3+1)=13(3+3)=39+133= \frac{26\sqrt{3}(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{26\sqrt{3}(\sqrt{3}+1)}{2} = 13\sqrt{3}(\sqrt{3}+1) = 13(3 + \sqrt{3}) = 39 + 13\sqrt{3}

Answer: S6=39+133S_6 = 39 + 13\sqrt{3}

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3(ii)Find the sum to 20 terms of the G.P.: 0.15+0.015+0.0015+…0.15 + 0.015 + 0.0015 + \ldotsShow solution
Given: a=0.15a = 0.15, r=0.1r = 0.1, n=20n = 20.

S20=a(1−r20)1−r=0.15(1−(0.1)20)1−0.1=0.150.9[1−(0.1)20]=16[1−(0.1)20]S_{20} = \frac{a(1 - r^{20})}{1 - r} = \frac{0.15(1 - (0.1)^{20})}{1 - 0.1} = \frac{0.15}{0.9}\left[1 - (0.1)^{20}\right] = \frac{1}{6}\left[1 - (0.1)^{20}\right]

Answer: S20=16[1−(0.1)20]S_{20} = \dfrac{1}{6}\left[1 - (0.1)^{20}\right]

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4Evaluate ∑k=110(3+2k)\displaystyle\sum_{k=1}^{10}(3 + 2^k).Show solution
∑k=110(3+2k)=∑k=1103+∑k=1102k\sum_{k=1}^{10}(3 + 2^k) = \sum_{k=1}^{10} 3 + \sum_{k=1}^{10} 2^k

=3×10+(21+22+…+210)= 3 \times 10 + (2^1 + 2^2 + \ldots + 2^{10})

=30+2(210−1)2−1=30+2(1024−1)=30+2046=2076= 30 + \frac{2(2^{10}-1)}{2-1} = 30 + 2(1024 - 1) = 30 + 2046 = 2076

Alternatively expressed: 28+211=28+2048=207628 + 2^{11} = 28 + 2048 = 2076.

Answer: ∑k=110(3+2k)=2076\displaystyle\sum_{k=1}^{10}(3+2^k) = 2076

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5The sum of the first two terms of a G.P. is 36 and the product of the first term and the third term is 9 times the second term. Find the sum of first 8 terms.Show solution
Given: a1+a2=36a_1 + a_2 = 36 and a1⋅a3=9a2a_1 \cdot a_3 = 9a_2.

Let first term =a= a, common ratio =r= r.

Condition 2: a⋅ar2=9⋅ara \cdot ar^2 = 9 \cdot ar
a2r2=9ar⇒ar=9⇒a2=9a^2r^2 = 9ar \Rightarrow ar = 9 \Rightarrow a_2 = 9

Condition 1: a+9=36⇒a=27a + 9 = 36 \Rightarrow a = 27

r=927=13r = \frac{9}{27} = \frac{1}{3}

Sum of 8 terms:
S8=27(1−(13)8)1−13=27(1−16561)23=27×6560656123=27×6560×36561×2=6560162=328081S_8 = \frac{27\left(1 - \left(\frac{1}{3}\right)^8\right)}{1 - \frac{1}{3}} = \frac{27\left(1 - \frac{1}{6561}\right)}{\frac{2}{3}} = \frac{27 \times \frac{6560}{6561}}{\frac{2}{3}} = \frac{27 \times 6560 \times 3}{6561 \times 2} = \frac{6560}{162} = \frac{3280}{81}

Answer: S8=328081S_8 = \dfrac{3280}{81}

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6Find the sum to nn terms of the sequence: 7,77,777,7777,…7, 77, 777, 7777, \ldotsShow solution
Given: Sn=7+77+777+…S_n = 7 + 77 + 777 + \ldots to nn terms.

Sn=7(1+11+111+… to n terms)S_n = 7(1 + 11 + 111 + \ldots \text{ to } n \text{ terms})
=79(9+99+999+…)= \frac{7}{9}(9 + 99 + 999 + \ldots)
=79[(10−1)+(102−1)+(103−1)+…+(10n−1)]= \frac{7}{9}[(10-1) + (10^2-1) + (10^3-1) + \ldots + (10^n-1)]
=79[(10+102+…+10n)−n]= \frac{7}{9}\left[(10 + 10^2 + \ldots + 10^n) - n\right]
=79[10(10n−1)9−n]= \frac{7}{9}\left[\frac{10(10^n - 1)}{9} - n\right]
=70(10n−1)81−7n9= \frac{70(10^n - 1)}{81} - \frac{7n}{9}

Answer: Sn=70(10n−1)81−7n9S_n = \dfrac{70(10^n - 1)}{81} - \dfrac{7n}{9}

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7The sum of first three terms of a G.P. is 3910\dfrac{39}{10} and their product is 1. Find the common ratio and the terms.Show solution
Let the three terms be ar,a,ar\dfrac{a}{r}, a, ar.

Product: ar⋅a⋅ar=a3=1⇒a=1\dfrac{a}{r} \cdot a \cdot ar = a^3 = 1 \Rightarrow a = 1

Sum: 1r+1+r=3910\dfrac{1}{r} + 1 + r = \dfrac{39}{10}

1+r+r2r=3910\frac{1 + r + r^2}{r} = \frac{39}{10}
10(1+r+r2)=39r10(1 + r + r^2) = 39r
10r2−29r+10=010r^2 - 29r + 10 = 0
(2r−5)(5r−2)=0(2r - 5)(5r - 2) = 0
r=52orr=25r = \frac{5}{2} \quad \text{or} \quad r = \frac{2}{5}

Terms:
- If r=52r = \dfrac{5}{2}: terms are 25,1,52\dfrac{2}{5}, 1, \dfrac{5}{2}
- If r=25r = \dfrac{2}{5}: terms are 52,1,25\dfrac{5}{2}, 1, \dfrac{2}{5}

Answer: r=52r = \dfrac{5}{2} or 25\dfrac{2}{5}; terms are 25,1,52\dfrac{2}{5}, 1, \dfrac{5}{2} (or in reverse order).

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8Find four numbers forming a G.P. in which the third term is greater than the first term by 9, and the second term is greater than the 4th by 18.Show solution
Let the four terms be a,ar,ar2,ar3a, ar, ar^2, ar^3.

Condition 1: ar2−a=9⇒a(r2−1)=9…(1)ar^2 - a = 9 \Rightarrow a(r^2 - 1) = 9 \quad \ldots(1)

Condition 2: ar−ar3=18⇒ar(1−r2)=18…(2)ar - ar^3 = 18 \Rightarrow ar(1 - r^2) = 18 \quad \ldots(2)

Divide (2) by (1):
ar(1−r2)a(r2−1)=189\frac{ar(1-r^2)}{a(r^2-1)} = \frac{18}{9}
−ar(r2−1)a(r2−1)=2\frac{-ar(r^2-1)}{a(r^2-1)} = 2
−r=2⇒r=−2-r = 2 \Rightarrow r = -2

From (1): a(4−1)=9⇒a=3a(4-1) = 9 \Rightarrow a = 3

Four terms: 3, 3(−2), 3(−2)2, 3(−2)3=3,−6,12,−243,\ 3(-2),\ 3(-2)^2,\ 3(-2)^3 = 3, -6, 12, -24

Answer: The four numbers are 3,−6,12,−243, -6, 12, -24.

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9Insert 6 geometric means between 27 and 181\dfrac{1}{81}.Show solution
Given: 6 G.M.s between 27 and 181\dfrac{1}{81}. Total terms =8= 8.

So: a1=27a_1 = 27, a8=181a_8 = \dfrac{1}{81}, r=?r = ?

a8=a1r7⇒181=27r7a_8 = a_1 r^7 \Rightarrow \frac{1}{81} = 27 r^7
r7=181×27=12187=137=(13)7r^7 = \frac{1}{81 \times 27} = \frac{1}{2187} = \frac{1}{3^7} = \left(\frac{1}{3}\right)^7
r=13r = \frac{1}{3}

Six G.M.s:
G1=27×13=9,G2=3,G3=1,G4=13,G5=19,G6=127G_1 = 27 \times \frac{1}{3} = 9, \quad G_2 = 3, \quad G_3 = 1, \quad G_4 = \frac{1}{3}, \quad G_5 = \frac{1}{9}, \quad G_6 = \frac{1}{27}

Answer: The 6 geometric means are 9,3,1,13,19,1279, 3, 1, \dfrac{1}{3}, \dfrac{1}{9}, \dfrac{1}{27}.

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10If the AM of two unequal positive real numbers aa and bb (a>ba > b) is twice as much as their GM, show that a:b=(2+3):(2−3)a : b = (2+\sqrt{3}) : (2-\sqrt{3}).Show solution
Given: A=2GA = 2G, where A=a+b2A = \dfrac{a+b}{2} and G=abG = \sqrt{ab}.

a+b2=2ab\frac{a+b}{2} = 2\sqrt{ab}
a+b=4ab…(1)a + b = 4\sqrt{ab} \quad \ldots (1)

Let ab=k\dfrac{a}{b} = k. Then:
a+bab=4\frac{a+b}{\sqrt{ab}} = 4
a/b+b/a1=4⇒k+1k=4\frac{\sqrt{a/b} + \sqrt{b/a}}{1} = 4 \Rightarrow \sqrt{k} + \frac{1}{\sqrt{k}} = 4

Let k=t\sqrt{k} = t:
t+1t=4⇒t2−4t+1=0t + \frac{1}{t} = 4 \Rightarrow t^2 - 4t + 1 = 0
t=4±122=2±3t = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}

Since a>ba > b, k>1k > 1, so t=2+3t = 2 + \sqrt{3}.

ab=2+3⇒ab=(2+3)2=7+43\sqrt{\frac{a}{b}} = 2 + \sqrt{3} \Rightarrow \frac{a}{b} = (2+\sqrt{3})^2 = 7 + 4\sqrt{3}

Also: ab=(2+3)21=(2+3)(2+3)(2+3)(2−3)⋅(2−3)\dfrac{a}{b} = \dfrac{(2+\sqrt{3})^2}{1} = \dfrac{(2+\sqrt{3})(2+\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})} \cdot (2-\sqrt{3})

More directly: ab=2+32−3\dfrac{a}{b} = \dfrac{2+\sqrt{3}}{2-\sqrt{3}} (rationalising: (2+3)(2+3)=7+43(2+\sqrt{3})(2+\sqrt{3}) = 7+4\sqrt{3} and (2−3)(2+3)=1(2-\sqrt{3})(2+\sqrt{3})=1, so 2+32−3=(2+3)2=7+43\dfrac{2+\sqrt{3}}{2-\sqrt{3}} = (2+\sqrt{3})^2 = 7+4\sqrt{3} ✓).

∴a:b=(2+3):(2−3)■\therefore a : b = (2+\sqrt{3}) : (2-\sqrt{3}) \quad \blacksquare

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11If a,b,c,da, b, c, d are in G.P., show that:
(i) a2+b2, b2+c2, c2+d2a^2+b^2,\ b^2+c^2,\ c^2+d^2 are in G.P.
(ii) 1a2+b2, 1b2+c2, 1c2+d2\dfrac{1}{a^2+b^2},\ \dfrac{1}{b^2+c^2},\ \dfrac{1}{c^2+d^2} are in G.P.Show solution
Given: a,b,c,da, b, c, d are in G.P. with common ratio rr, so b=arb = ar, c=ar2c = ar^2, d=ar3d = ar^3.

(i) Let P=a2+b2P = a^2+b^2, Q=b2+c2Q = b^2+c^2, R=c2+d2R = c^2+d^2.

P=a2+a2r2=a2(1+r2)P = a^2 + a^2r^2 = a^2(1+r^2)
Q=a2r2+a2r4=a2r2(1+r2)Q = a^2r^2 + a^2r^4 = a^2r^2(1+r^2)
R=a2r4+a2r6=a2r4(1+r2)R = a^2r^4 + a^2r^6 = a^2r^4(1+r^2)

QP=r2,RQ=r2\frac{Q}{P} = r^2, \quad \frac{R}{Q} = r^2

Since the ratio is constant, P,Q,RP, Q, R are in G.P. ■\blacksquare

(ii) From (i), P,Q,RP, Q, R are in G.P., so Q2=PRQ^2 = PR.

1Q2=1PR⇒1Q⋅1Q=1P⋅1R\frac{1}{Q^2} = \frac{1}{PR} \Rightarrow \frac{1}{Q} \cdot \frac{1}{Q} = \frac{1}{P} \cdot \frac{1}{R}

This means 1P,1Q,1R\dfrac{1}{P}, \dfrac{1}{Q}, \dfrac{1}{R} are in G.P. ■\blacksquare

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12Let SS be the sum, PP the product and RR the sum of reciprocals of nn terms of a G.P. Prove that P2Rn=SnP^2 R^n = S^n.
13What will Rs. 5000 amount to in 10 years after it is deposited in a bank which pays annual interest of 8% compounded annually?
14If the first and the nthn^{\text{th}} term of a G.P. are aa and bb respectively, and if PP is the product of nn terms, prove that P2=(ab)nP^2 = (ab)^n.
15A certain type of bacteria doubles its population every 20 minutes. Assuming no bacteria die, how many bacteria will be there after 3 hours if there are 1 million bacteria at present?
16One side of an equilateral triangle is 24 cm. The midpoints of its sides are joined to form another triangle whose midpoints are joined to form yet another triangle and so on. This process continues indefinitely. Find the sum of the perimeters of all the triangles.
17After striking a floor, a certain ball rebounds 45\dfrac{4}{5}th of the height from which it has fallen. If the ball is dropped from a height of 240 cm, find the total distance the ball travels before coming to rest.
18An object decelerates such that it travels 60 m during the first second, 20 m during the second and 203\dfrac{20}{3} m during the third second. Determine the total distance the object travels before coming to rest.
19Suppose a person mails a letter to five of his friends. He asks each one of them to mail it further to five additional friends with instruction that they move the chain further. Assuming the chain is not broken and no person receives the mail more than once, determine the amount spent on postage when the 8th set of letters is mailed, if cost of postage of each letter is 50 paisa.
20Due to reduced taxes an individual has an extra Rs. 30,000 in spendable income. If we assume that an individual spends 70% of this on consumer goods and the producers of these goods in turn spend 70% on consumer goods and this process continues indefinitely. What is the total amount spent on consumer goods?
21A machine depreciates in value by one-fifth each year. If the machine is now worth Rs. 51,000, how much will it be worth 3 years from now?
22The sum of an infinite G.P. is 3 and the sum of the squares of its terms is also 3. Then its first term and common ratio are:
(i) 1,121, \dfrac{1}{2} (ii) 12,32\dfrac{1}{2}, \dfrac{3}{2} (iii) 32,12\dfrac{3}{2}, \dfrac{1}{2} (iv) 1,141, \dfrac{1}{4}
23An antique's present worth is Rs. 9000. If its value appreciates at the rate of 10% per year, its worth 3 years from now is:
(i) Rs. 6561 (ii) Rs. 10,890 (iii) Rs. 11,979 (iv) Rs. 12,000
24Venessa invests Rs. 5000 in a bond that pays 6% interest compounded semi-annually. The value of the bond in rupees after 5 years is:
(i) 5000(1.06)55000(1.06)^5 (ii) 5000(1.03)55000(1.03)^5 (iii) 5000(1.06)105000(1.06)^{10} (iv) 5000(1.03)105000(1.03)^{10}

Practice Questions — Multiple Choice Questions

1Find the sum of the G.P.: 810,8100,81000,810000,…\dfrac{8}{10}, \dfrac{8}{100}, \dfrac{8}{1000}, \dfrac{8}{10000}, \ldots to nn terms.
(a) −89(110n−1)\dfrac{-8}{9}\left(\dfrac{1}{10^n}-1\right) (b) 881(110n−1)\dfrac{8}{81}\left(\dfrac{1}{10^n}-1\right) (c) 98(110n−1)\dfrac{9}{8}\left(\dfrac{1}{10^n}-1\right) (d) 890(110n−1)\dfrac{8}{90}\left(\dfrac{1}{10^n}-1\right)
2If the first term of a GP is 5 and common ratio is (−5)(-5), then which term is 3125?
(a) 6th6^{\text{th}} (b) 8th8^{\text{th}} (c) 5th5^{\text{th}} (d) 4th4^{\text{th}}
3Which number should be added to the numbers 3, 8, 13 to make the resulting numbers a G.P.?
(a) 4 (b) 2 (c) 5 (d) −2-2
4If the third term of a G.P. is 6, then the product of its first 5 terms is:
(a) 565^6 (b) 656^5 (c) 525^2 (d) 626^2
5If aa, bb and cc are in A.P. as well as in G.P., then which of the following is true?
(a) a=b≠ca = b \neq c (b) a≠b≠ca \neq b \neq c (c) a=b=ca = b = c (d) a≠b=ca \neq b = c

Practice Questions — Problems

1Your friend has invested in a 'Grow Your Money Scheme' that promises to return Rs. 11,000 after a year if you invest Rs. 1,000 at the rate of 10% compounded annually. Would you be willing to invest in this scheme? Explain.
2Write the first four terms of a geometric series for which Sn=39,360S_n = 39,360 and r=3r = 3.
3Using Geometric series, write 0.175175175…0.175175175\ldots in fraction.
4In a mock test, Rohan and Shweta solved: Find the 10th term of the Geometric series 9,3,1,…9, 3, 1, \ldots
(a) Who is correct? Explain your reasoning.
(b) Can you guess the correct answer without solving? If yes, what argument would you use?
5On the first day, a music video of Arijit Singh posted online got 120 views in Delhi. The number of viewership increases by 5% per day. How many total views did the video get over the course of the first 29 days? Express your answer in exponential form.
6Harry traced his family back for 15 generations starting with his parents. How many ancestors did he have in total?
7You and your sibling decide to ask for a raise in pocket money. Your Dad gives both of you a choice: Rs. 1000 at once, or Rs. 2 on day one, Rs. 4 on day two, doubling each day for 12 days. You opted for Rs. 1000 at once; your brother opted for the doubling scheme. Which one made a better decision and why?
8If ax=by=cza^x = b^y = c^z such that aa, bb and cc are in G.P. and xx, yy and zz are unequal positive integers, then show that 2y=1x+1z\dfrac{2}{y} = \dfrac{1}{x} + \dfrac{1}{z}.
9A person sends a fake news on WhatsApp to 4 of his friends on Monday. Each of those friends forwards it to 4 of their friends on Tuesday, and so on for a week. Find how many people have received the fake news on WhatsApp till then?
10Three positive numbers form an increasing G.P. If the middle term of the series is doubled, then the new numbers are in A.P. Find the common ratio of the G.P.
11Priyanka invested Rs. 1300 in an account that pays 4% interest compounded annually. Assuming no deposits or withdrawals are made, find how much money she would have in the account 6 years after her initial investment.
12A financial analyst is analyzing a company. A dividend of Rs. 500 has just been paid. Dividends will grow by 20% per year for the next 3 years, followed by annual growth of 10% per year for 2 years.
(a) Complete the table for Years 1–5.
(b) Calculate the total dividend for the next five years.
13(a) Rs. 1,000 is invested for three years at 6% per annum compounded semi-annually. Calculate the total return after three years.
(b) What would the answer be if the interest was compounded annually?
(c) What can you infer about the frequency of compounding and the size of the total return?
14From the graph, what conclusion can be drawn regarding the frequency of compounding? (Graph not visible in OCR)
15A small country emits 130,100 kilotons of carbon dioxide per year. In the first year it will keep emissions at 130,000 kilotons and emissions will decrease 3.1% in each of the next two years. How many kilotons of carbon dioxide would the country emit over the course of the 3-year period?
17A stock begins to pay dividends with the first dividend, one year from now, expected to be Rs. 100. Each year the dividend is 10% larger than the previous year's dividend. In what year will the dividend paid be larger than Rs. 1000? (Use concept of logarithm)

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