Exercise 9.1
Q1: as = s (s + 3) is the sth term of a sequence. Find the starting five terms of the sequence.
Answer:
as = s (s + 3)
Putting s = 1, 2, 3, 4 and 5 respectively, in as = s (s + 3)
a1 = 1 (1 + 3) = 4
a2 = 2 (2 + 3) = 10
a3 = 3 (3 + 3) = 18
a4 = 4 (4 + 3) = 28
a5 = 5 (5 + 3) = 40
The starting five terms of the sequence are 4, 10, 18, 28 and 40
Q2: as = ss+2 is the sth term of a sequence. Find the starting five terms of the sequence.
Answer:
as = ss+2
Putting s = 1, 2, 3, 4 and 5 respectively, in ss+2
a1 = 11+2 = 13
a2 = 22+2 = 24
a3 = 33+2 = 35
a4 = 44+2 = 46
a5 = 55+2 = 57
The starting five terms of the sequence are 12,23,34,45and56
Q3: as = 5s is the sth term of a sequence. Find the starting five terms of the sequence.
Answer:
as = 5s
Putting s = 1, 2, 3, 4 and 5 respectively, in as = 5s
a1 = 51 = 5
a2 = 52 = 25
a3 = 53 = 125
a4 = 54 = 625
a5 = 55 = 3125
The starting five terms of the sequence are 5, 25, 125, 625 and 3125.
Q4: as = 2s+23 is the sth term of a sequence. Find the starting five terms of the sequence.
Answer:
as = 2s+23
Putting s = 1, 2, 3, 4 and 5 respectively, in 2s+23
a1 = 2.1+23=43
a2 = 2.2+23=63
a3 = 2.3+23=83
a4 = 2.4+23=103
a5 = 2.5+23=123
The starting five terms of the sequence are 43,63,83,103and123
Q5: as = (−1)s–1+3s+1 is the sth term of a sequence. Find the starting five terms of the sequence.
Answer:
as = (−1)s–1+3s+1
Putting s = 1, 2, 3, 4 and 5 respectively, in (−1)s–1+3s+1
a1 = (−1)1–1+31+1 = 1 + 9 = 10
a2 = (−1)2–1+32+1 = – 1 + 27 = 26
a3 = (−1)3–1+33+1 = 1 + 81 = 82
a4 = (−1)4–1+34+1 = – 1 + 243 = 242
a5 = (−1)5–1+35+1 = 1 + 729 = 730
The starting five terms of the sequence are 10, 26, 82, 242 and 730.
Q6: as = (−1)s–13s+1 is the sth term of a sequence. Find the starting five terms of the sequence.
Answer:
as = (−1)s–13s+1
Putting s = 1, 2, 3, 4 and 5 respectively, in (−1)s–13s+1
a1 = (−1)1–131+1=(−1)032=19
a2 = (−1)2–132+1=(−1)133=−127
a3 = (−1)3–133+1=(−1)234=181
a4 = (−1)4–134+1=(−1)335=−1243
a5 = (−1)5–135+1=(−1)436=1729
The starting five terms of the sequence are 19,−127,181,−1243,and1729
Q7: as = s + 3s is the sth term of a sequence. Obtain the 17th and 22nd term in the required sequence.
Answer:
as = s + 3s
Putting s = 17 and 22 respectively, in as = s + 3s
a17 = 17 + 317
a22 = 22 + 322
Q8: as = 9+ss2 is the sth term of a sequence. Obtain the 24th term in the required sequence.
Answer:
as = 9+ss2
Putting s = 24, in as = 9+ss2
a24 = 9+24242=33576
Q9: as = (−1)s×6s is the sth term of a sequence. Obtain the 11th and 12th term in the required sequence.
Answer:
as = (−1)s×6s
Putting s = 11 and 12, in as = (−1)s×6s
a11 = (−1)11×611=–611
a12 = (−1)12×612=612
Q10. as = s+32+s is the sth term of a sequence. Obtain the 21st and 82nd term in the required sequence.
Answer:
as = s+32+s
Putting s = 21 and 82, in as = s+32+s a21=21+32+21=2423 a82=82+32+82=8584
Q11. Find out the starting five terms and also the following series of the required sequence given below:
a1 = 2, as = 2 as – 1 for all s > 1
Answer:
a1 = 2, as = 2 as – 1 for all s > 1
a2 = 2 a1 = 2 x 2 = 4
a3 = 2 a2 = 2 x 4 = 8
a4 = 2 a3 = 2 x 8 = 16
a5 = 2 a4 = 2 x 16 = 32
The starting five terms of the sequence are 2, 4, 8, 16 and 32
The following series is 2 + 4 + 8 + 16 + 32 + …….
Q12. Find out the starting five terms and also the following series of the required sequence given below:
a1 = 2, as = as–1–12 for all s > 3
Answer:
a1 = 2, as = as–1–12 for all s > 3
a2–1–12=a1–12=2–12=12a3=a3–1–12=a2–12=12–12=–14a4=a4–1–12=a3–12=−14–12=–58a5=a5–1–12=a4–12=−58–12=–1316
The starting five terms of the sequence are 2,12,–14,–58and–1316
The following series is 2+12+(−14)+(−58)+(−1316)+……
Q13. Find out the starting five terms and also the following series of the required sequence given below:
Answer
a1 = a2 = 4, as = as – 1 + 1 for all s > 4
a3 = a 3 – 1 + 1 = a2 +1 = 4 + 1 = 5
a4 = a 4 – 1 + 1 = a3 + 1 = 5 + 1 = 6
a5 = a 5 – 1 + 1 = a4 + 1 = 6 + 1 = 7
The starting five terms of the sequence are 4, 4, 5, 6 and 7
The following series is 4 + 4 + 5 + 6 + 7 + ….
Q14. The sequence of Fibonacci is described as 1 = a1 = a2 and as = as – 1 + as – 2, n > 1. Obtain as+1as for n = 2, 4.
Answer:
1 = a1 = a2 and as = as – 1 + as – 2, n > 1.
a3 = a2 + a1 = 1 + 1 = 2
a4 = a3 + a2 = 2 + 1 = 3
a5 = a4 + a3 = 3 + 2 = 5
a6 = a5 + a4 = 5 + 3 = 8
For s = 2 and 4 respectively,
a2+1a2=a3a2=21=2 a4+1a4=a5a4=53
Exercise 9.2
Q1. Obtain the sum of all integers which are odd between 1 and 4001 including 1 and 4001.
Answer:
The integers which are odd between 1 and 2001 are 1, 3, 5, 7, 9, ……………. , 3997, 3999, 4001.
The sequence is in A.P form.
a = 1 [1st term]
Difference, d = 2
The standard equation of A.P,
a + (s – 1) d = 4001
1 + (s – 1) 2 = 4001
s = 2001
SS = s2[2a+(s–1)d]Ss=20012[2.1+(2001–1)2]=20012[2+4000]=20012×4002=2001×2001=4004001
Hence, the sum of all integers which are odd between 1 and 4001 is 4004001.
Q2. Obtain the sum of all natural numbers which are lying between 10 and 100, and are multiples of 5.
Answer:
All natural numbers which are lying between 10 and 100, and are multiples of 5, are 15, 20, 25, … , 95.
The sequence is in A.P form.
a = 15 [1st term]
Difference, d = 5
The standard equation of A.P,
a + (s – 1) d = 95
15 + (s – 1) 5 = 95
s = 17
SS = s2[2a+(s–1)d]Ss=172[2.15+(17–1)5]=172[30+80]=172×110=17×55=935
Hence, the sum of all natural numbers which are lying between 10 and 100, and are multiples of 5 is 935.
Q3. Prove that the 21st term in the sequence is – 118 provided that the sequence is in A.P form, the sum of the starting 5 terms is one – fourth of the next 5 terms and 2 is the first term.
Answer:
Given:
2 is the first term, a = 2
The sequence is in A.P form,
In A. P = 2, 2 + d, 2 + 2d, 2 + 3d ……
In A.P form, the sum of the starting 5 terms is one – fourth of the next 5 terms,
So, according to the given condition,
10 + 10d = (1 / 4) (10 + 35d)
40 + 40d = 10 + 35d
d = – 6
a21 = a + (21 – 1) d = 2 + 20 (- 6) = – 118
Hence proved
Q4. Obtain the number of terms in A.P which are needed to get – 25 from the sum of – 6, (- 11 / 2), – 5, ….
Answer:
Suppose,
In, A.P the sum of s terms = – 25
a = – 6 [1st term]
Difference, d = (- 11 / 2) + 6 = (1 / 2)
s2[2a+(s–1)d] −25=s2[2.(−6)+(n–1)12]–50=n[−12+n2–12]–50=n[−25+n2]–100=n(−25+n)n2–25n+100=0n2–5n–20n+100=0(n–5)(n–20)=0n=5or20
Q5. If m th term is 1 / n and nth term is 1 / m provided that the sequence is in A.P form, prove that the sum of the first mn terms is (1 / 2) (mn + 1), where m n.
Answer:
Given,
m th term is 1 / n and nth term is 1 / m
So, acc. to the above condition
m th term in A.P form = am = a + (m – 1) d = 1 / m ……. (1)
n th term in A.P form = an = a + (n – 1) d = 1 / n ……. (2)
(2) – (1), we get,
(n – 1) d – (m – 1) d = 1m–1n=n–mmn
(n – m) d = n–mmn
d = 1mn
Substituting the value of d in equation (1), we get,
a + (m – 1) 1mn = 1 / m
a = a=1n–1n+1mn=1mn Smn=mn2[2a+(mn–1)d]Smn=mn2[21mn+(mn–1)1mn]=1+12(mn–1)=12(mn+1)
Hence proved
Q6. Obtain the last term if the addition of some numbers in A.P. 25, 22, 19, is 116.
Answer:
Addition of s terms in A.P be 116
Here, first term a = 25, d = – 3
Ss=s2[2a+(s–1)d]116=s2[2.(25)+(s–1)(−3)]116=s[50–3s+3]232=s(53–3s)=53s–s23s2–53n+232=03s2–24n–29n+232=0(3s–29)(n–8)=0n=8orn=(293)
n = 8 is considered
a8 (last term) = a + (s – 1) d = 25 + 7 (- 3) = 25 – 21 = 4
Hence, the last term is 4
Q7. In A.P, obtain the total to s terms whose nth term is 5n + 1.
Answer:
Given,
In A.P, the total to s terms whose nth term is 5n + 1.
nth term = an = a + (n – 1) d
an = a + (n – 1) d = 5n + 1
a + nd – d = 5n + 1
By comparing the coefficient of n, we get
d = 5 and a – d = 1
a – 5 = 1
a = 6
Ss=s2[2a+(s–1)d]=s2[2.6+(s–1)d]=s2[12+(s–1)5]=s2[12+5s–5]=s2[5s+7].
Q8. Obtain the common difference of a sequence in A.P, when the total of s terms is (ms + ns2), where m and n are constants.
Answer:
Given,
As mentioned in the given condition,
Ss=s2[2a+(s–1)d] = (ms + ns2)
s2[2a+(s–1)d]=ms+ns2s2[2a+sd–d]=ms+ns2sa+s2d2–sd2=ms+ns2
Considering the coefficients of the of s2 , we get,
d2=n
d = 2n
Hence, in A.P the difference d = 2n
Q9. The ratio of the total of s terms of two arithmetic progressions are 5s + 4 : 9 + 6 . Obtain the ratio of 18th term.
Answer:
Suppose the first terms are a1 and a2 respectively, and the common differences be d1 and d2 of the first two consecutive arithmetic progressions respectively.
As mentioned in the given condition,
Totaladditionofstermsof1stA.PTotaladditionofstermsof2ndA.P=5s+49s+6s2[2a1+(s–1)d1]s2[2a2+(s–1)d2]=5s+49s+62a1+(s–1)d12a2+(s–1)d2=5s+49s+6Puttings=35in(1),weget2a1+34d12a2+34d2=5(35)+49(35)+6a1+17d1a2+17d2=179321
18th term = a18=182[2a+(18–1)d] 18thtermof1stA.P18thtermof2ndA.P=a18=182[2a+(18–1)d]=a1+17d1a2+17d2=179321
Hence, 179 : 321 is the required ratio of the 18th term.
Q10. In an A.P, the total of starting m terms is equal to the total of starting n terms. Obtain the total of starting (m + n) terms.
Answer:
m th term = Sm = m2[2a+(m–1)d]
n th term = Sn = n2[2a+(n–1)d]
As mentioned in the given condition,
m2[2a+(m–1)d]=n2[2a+(n–1)d]m[2a+(m–1)d]=n[2a+(n–1)d]2am+(m–1)md=2an+(n–1)nd2a(m–n)+d[m(m–1)–n(n–1)]=02a(m–n)+d[m2–m–n2+n]=02a(m–n)+d[(m+n)(m–n)–(m–n)]=02a(m–n)+d[(m–n)(m+n–1)]=02a+d(m+n–1)=0d=−2am+n–1Sm+n=m+n2[2a+(m+n–1)d]Sm+n=m+n2[2a+(m+n–1)−2am+n–1]Sm+n=m+n2[2a–2a]Sm+n=0
Hence, the total of starting (m + n) terms is 0.
Q11. . In an A.P, the total of starting m, n and o terms are p, q and r. Show that the pm(n–o)+qn(m–o)+ro(m–n)=0.
Answer:
Given,
In an A.P, the total of starting m, n and o terms are p, q and r.
As mentioned in the given condition,
Sm=m2[2a1+(m–1)d]=p2a1+(m–1)d=2pm……(i)Sn=n2[2a1+(n–1)d]=q2a1+(n–1)d=2qn……(ii)So=o2[2a1+(o–1)d]=r2a1+(o–1)d=2ro……(iii)Subtract(i)−(ii),weget(m–1)d–(n–1)d=2pm–2qnd(m–1–n+1)=2pn–2qmmnd(m–n)=2(pn–qm)mnd=2(pn–qm)mn(m–n)….(4) Subtract(ii)−(iii),weget(n–1)d–(o–1)d=2qn–2rod(n–1–o+1)=2qo–2rnnod(n–o)=2(qo–rn)nod=2(qo–rn)no(n–o)….(5)
Now, equating the values of d in equation (4) and (5)
2(pn–qm)mn(m–n)=2(qo–rn)no(n–o)no(n–o)(pn–qm)=mn(m–n)(qo–rn)o(n–o)(pn–qm)=m(m–n)(qo–rn)(n–o)(pno–qmo)=(m–n)(mqo–mrn)Multiplyboththesidesby1mno,weget,(pm–qn)(n–o)=(qn–ro)(m–n)pm(n–o)–qn(n–o)=qn(m–n)–ro(m–n)pm(n–o)+qn(n–o+m–n)+ro(m–n)=0pm(n–o)+qn(m–o)+ro(m–n)=0
Hence proved
Q12. The ratio of the total of r and s terms of two arithmetic progressions is r2 : s2. Show that the ratio of r th and s th term is (2r – 1) : (2s – 1).
Answer:
Suppose the first term is a, and the common difference be d of the arithmetic progressions respectively.
As mentioned in the given condition,
SumofrtermsSumofsterms=r2s2r2[2a+(r–1)d]s2[2a+(s–1)d]=r2s22a+(r–1)d2a+(s–1)d=rs…..(i)Substitutingr=2r–1ands=2s–1in(i),weget,2a+(2r–1–1)d2a+(2s–1–1)d=2r–12s–12a+(2r–2)d2a+(2s–2)d=2r–12s–1a+(r–1)da+(s–1)d=2r–12s–1….(ii)rthtermofA.PsthtermofA.P=a+(r–1)da+(s–1)d=2r–12s–1rthtermofA.PsthtermofA.P=2r–12s–1
Hence proved
Q13. In an A.P, the total of n terms is 3p 2 + 5p and its rth term is 164. Obtain the value of r.
Answer:
Given,
Sr = r2[2a+(r–1)d] = 164 …… (i)
Suppose the first term is a, and the common difference be d of the arithmetic progressions respectively.
As mentioned in the given condition,
p th term = Sp = p2[2a+(p–1)d] = 3 p 2 + 5p
As mentioned in the given condition,
pa+d2p2–d2p=3p2+5pd2p2+(a–d2)p=3p2+5pEquating,thecoefficientsofp2onboththesides,wegetd2=3d=6Equating,thecoefficientsofponboththesides,wegeta–d2=5a–3=5a=8
From equation (i), we get
8 + (r – 1) 6 = 164
(r – 1) 6 = 164 – 8
(r – 1) 6 = 156
(r – 1) = 26
r = 7
Q14. Place five numbers between 8 and 26 in a way such that the sequence results in an A.P form.
Answer:
Suppose Q1, Q2, Q 3, Q 4 and Q 5 be the five required numbers.
First term, a = 8,
Last term, p = 26
s = 7,
p = a + (s – 1) d
26 = 8 + (7 – 1) d
6 d = 18
d = 3
Q1 = a + d = 8 + 3 = 11
Q2 = a + 2 d = 8 + 2.3 = 14
Q 3 = a + 3 d = 8 + 3.3 = 17
Q 4 = a + 4 d = 8 + 3.4 = 20
Q 5 = a + 5 d = 8 + 3. 5 = 23
Hence, Q1, Q2, Q 3, Q 4 and Q 5 is equal to 11, 14, 17, 20 and 23 are the required numbers in an A.P
Q15. Suppose in an A.M pm+qmpm–1+qm–1 lies between p and q terms, then obtain the value of m.
Answer:
A.M of p and q = p+q2
As mentioned in the given condition,
p+q2=pm+qmpm–1+qm–1(p+q)(pm–1+qm–1)=2(pm+qm)pm+pqm–1+qpm–1+qm=2pm+2qmpqm–1+qpm–1=pm+qmqm–1(p–q)=pm–1(p–q)qm–1=pm–1(pq)m–1=1=(pq)0m–1=0m=1
Q16. Insert n numbers between 1 and 31 in a way such that the sequence results in an A.P. The ratio is 5 : 9 of the 7th and the (n – 1)th term. Find the value of n.
Answer:
Suppose Q1, Q2, Q 3, Q 4.….. Q m, be the required n numbers.
First term, a = 1,
Last term, p = 31
s = n + 2,
p = a + (s – 1) d
31 = 1 + (n + 2 – 1) d
30 = (n + 1) d
d = 30(n+1)
Q1 = a + d
Q2 = a + 2 d
Q 3 = a + 3 d
Q 4 = a + 4 d …….
Q 7 = a + 7 d
Q n – 1 = a + (n – 1) d
As mentioned in the given condition,
a+7da+(n–1)d=591+7(30n+1)1+(n–1)(30n+1)=59n+1+7(30)n+1+30(n–1)=59n+21131m–29=599n+1899=155n–145155n–9n=1899+145146n=2044n=14
Hence, the value of n = 14
Q17. A boy will be repaying his loans and his first installment is Rs 100. What is the amount should he pay in the 30th installment if he increases the installment by Rs 5 every month?
Answer:
Given,
His first installment is Rs 100
His 2nd installment is Rs 105 and third installment is Rs 110 and so on
The sequence of money paid by the boy is in an A.P form every month is
100, 105, 110, 115 and so on …….
a = 100 [First term]
Common difference, d = 5
a 30 = a + (30 – 1) d
= 100 + 29 (5)
= 245
Hence, the amount should be paid by him in the 30th installment is Rs 245.
Q18. In a polygon, 120o is the smallest angle and 5o is the difference between consecutive angles on the interior side. Obtain the number of sides the polygon has.
Answer:
Given,
120o is the smallest angle i.e., first term, a = 120
And 5o is the difference, i.e., d = 5
Total of all angles of a polygon having s sides = 180o (s – 2)
Ss=180∘(s–2)s2[2a+(s–1)d=180∘(s–2)s[240+(s–1)5=360(s–2)240s+s(s–1)5=360(s–2)240s+5s2–5s=360s–7205s2+235s–360s+720=05s2–125s+720=0s2–25s+144=0s2–16s–9s+144=0s(s–16)–9(s–16)=0(s–16)(s–9)=0s=16ors=9
Exercise 9.3
Q1. Find 20th and nth term for the G.P 52,54,58,....
Soln:
Given G.P = 52,54,58,....
Here, first term = a = 52
Common ratio = r = 5452=12 a20=ar20–1=52(12)19=5(2)(2)19=5(2)20 an=arn–1=52(12)n–1=5(2)(2)n–1=5(2)n
Q2. Find 12th term for the G.P that has 8th term 192 and common ratio of 2.
Soln:
Given,
Common ratio = r = 2
Assume = first term = a
a8=ar8–1=ar7⇒ar7=192a(2)7=(2)6(3) ⇒a=(2)6×3(2)7=32 a12=ar12–1=(32)(2)11=(3)(2)10=3072
Q3. If p is the 5th, q is the 8th and s is the 11th term of a G.P. Prove that q2 = ps
Soln:
Let the first term be ‘a’ and the common ratio be r for the G.P.
Given condition
a5=ar5–1=ar4=p....(1) a8=ar8–1=ar7=q....(2) a11=ar11–1=ar10=s....(3)
Eqn (2) divided by Eqn (1), we have
ar7ar7=qp r3=qp . . . . . (4)
Eqn (3) divided by Eqn (2), we have
ar10ar7=sq ⇒r3=sq . . . . . (5)
From eqn (4) and eqn (5), we get
qp=sq ⇒q2=ps
Hence proved
Q4. If first term of a G.P is -3 and the 4th term is square of the 2nd term. Find the 7th term.
Soln:
Let the first term be ‘a’ and common ratio be ‘r’.
Therefore, a = -3
We know that, an = arn – 1
a4 = ar3 = (-3)r3
a2 = ar1 = (-3)r
Given condition
(-3)r3 = [(-3) r]2
⇒−3r3=9r2⇒r=–3a7=ar7–1=a
r6 = (-3) (-3)6 = – (3)7 = -2187
Hence, the 7th term is -2187.
Q5. Which term of
(a) 2,22–√,4,....is128?
(b) 3–√,3,33–√....is729?
(c) 13,19,127,....is119683?
Soln:
(a) Given sequence = 2,22–√,4,....
Here,
First term = a = 2
Common ratio = r = 22√2=2–√
Assume nth term = 128
an=arn–1 ⇒(2)(2–√)n–1=128 ⇒(2)(2)n–12=(2)7 ⇒(2)n–12+1=(2)7 n–12+1=7 ⇒n–12=6 ⇒n–1=12 ⇒n=13
Hence, 128 is the 13th term
(b) Given sequence = 3–√,3,33–√....
Here,
First term = a = 3–√
Common ratio = r = 33√=3–√
Assume nth term = 729
an=arn–1 arn–1=729 ⇒(3–√)(3–√)n–1=729 ⇒(3)12(3)n–12=(3)6 ⇒(3)12+n–12=(3)6
Therefore,
12+fracn–12=6 ⇒1+n–12=6 ⇒n=12
Hence, 729 is the 12th term of the sequence.
(c) Given sequence = 13,19,127,....
Here,
First term = a = 13
Common term = r = 19÷13=13
Assume the nth term be 119683 an=arn–1 arn–1=119683 ⇒(13)(13)n–1=119683 ⇒(13)n=(13)9
n = 9
Hence, 119683 is the 9th term of the sequence.
Q6. If 27,x,−72 are in G.P then find the value of x?
Soln:
Given sequence = 27,x,−72
Common ratio = x−27=−7x2
Also, −72x=−72x −7x2=−72x ⇒x2=−2×7−2×7=1 ⇒x=1–√ ⇒x=±1
Hence, the sequence is in G.P if x=±1
Q7. The sum of the first 20 terms of the G.P 0.15, 0.015, 0.0015 . . . .
Soln:
Given G.P = 0.15, 0.015, 0.0015 . . . .
Here,
First term = a = 0.15
Common ration = r = 0.0150.15=0.1 Sn=a(1–rn)1–r =0.15[1–(0.1)20]1–0.1 =0.150.9[1–(0.1)20] =1590[1–(0.1)20] =16[1–(0.1)20]
Q8. 7–√,21−−√,37–√,.... is a G.P series. Find sum till the nth term.
Soln:
Given series = 7–√,21−−√,37–√,....
Here,
First term = a = 7–√
Common ratio = r = 21√7=3–√ Sn=a(1–rn)1–r =7√[1–(3√)n]1–3√ =7√[1–(3√)n]1–3√×1+3√1+3√ =7√(3√+1)[1–(3√)n]1–3 =–7√(3√+1)[1–(3√)n]2
Q9. What is the sum of the nth term of the G.P series
1, -a, a2, -a3 . . . . (if a ≠ -1)?
Soln:
Given series = 1, -a, a2, -a3 . . .
Here,
The first term of the G.P = a1 = 1
Common ratio of the G.P = r = -a
Sn=a1(1–rn)1–r Sn=1[1–(−a)n]1–(−a)=[1–(−a)n]1+a
Q10. What is the sum of the nth term of the G.P series
x3, x5, x7 . . . . (if x ≠ ± 1)
Soln:
Given series = x3, x5, x7 . . . .
Here,
First term = a = x3
Common ratio = r = x2
Sn=a(1–rn)1–r=x3[1–(x2)n]1–x2=x3(1–x2n)1–x2
Q11. Evaluate ∑11k=1(2+3k)
Soln:
∑11k=1(2+3k)=∑11k=1(2)+∑11k=13k=2(11)+∑11k=13k=22+∑11k=13k.....(1) ∑11k=13k=31+32+33+....+311
The above sequence is in a G.P 3, 32, 33, . . .
Sn=a(rn–1)r–1 ⇒Sn=3[(3)11–1]3–1 ⇒Sn=32(311–1) ∑11k=13k=32(311–1) ….(2)
Substituting eqn (2) in eqn (1), we get
∑11k=1(2+3k)=22+32(311–1)
Q12. Sum of 1st 3 terms in a G.P is 3910 and the product of the terms is 1. What is the the terms and common ratio of the sequence?
Soln:
Assume ar,a,ar be the 1st 3 terms in the G.P
ar+a+ar=3910 . . . . (1)
(ar)(a)(ar)=1 . . . . (2)
From eqn (2), we get a3 = 1
a = 1
Substituting (a = 1) in eqn (1), we get
1r+1+r=3910 ⇒1+r+r2=3910r ⇒10+10r+10r2−39r=0 ⇒10r2–29r+10=0 ⇒10r2–25r–4r+10=0 ]⇒5r(2r–5)−2(2r–5)=0 ]⇒(5r–2)(2r–5)=0 ]⇒r=25or52
Hence,
The terms are 52,1and25
Q13. For the G.P 3, (3)2, (3)3 …. sum of how many terms is 120?
Soln:
Given G.P = 3, (3)2, (3)3 ….
Let n number of terms are required for obtaing the sum of 120
Sn=a(1–r)n1–r
Here,
First term = a = 3
Common ratio = r = 3
Sn=120=3(3n–1)3–1 ⇒120=3(3n–1)2 ⇒120×23=3n–1 ⇒3n–1=80 ⇒3n=81 ⇒3n=34
n = 4
Hence, 4 terms are required to get the sum of 120
Q14. In a G.P sum of the first 3 terms of are 16 where as sum for the next 3 terms are 128.
Find common ratio, first term and sum of n terms for the G.P
Soln:
Assume G.P = a, ar, a(r)2, a(r)3, . . .
Given condition =
a + a(r)2 + ar + = 16 and a(r)5 + a(r)4 + a(r)3 = 128
a (1 + (r)2 + r) = 16 …… (1)
a(r)3(1+ (r)2 + r) = 128 …….. (2)
eqn (2) divided by eqn (1), we get
ar3(1+r+r2)a(1+r+r2)=12816
r3 = 8
r = 2
substituting r = 2 in eqn(1), we get
a(1 + 2 + 4) = 16
a(7) = 16
a=167 Sn=167(2n–1)2–1=167(2n–1)
Q15. Given first term ‘a’ = 729 and the 7th term = 64, then determine S7.
Soln:
Given a = 729 , a7 = 64
Assume common ratio = r
We know that,
an = a rn – 1
a7 = a r7 – 1 = (729) (r)6
64 = 729 (r)6
r6=64729 r6=(23)6 r=23
Also, we know that
Sn=a(1–rn)1–r S7=729[1–(23)7]1–23 =3×729[1–(23)7] =(3)7[(3)7–(2)7(3)7] =(3)7–(2)7
= 2187 – 128
= 2059
Q16. The sum of first 2 terms of a G.P is -4 and 5th term is 4 times that of third term. Find the G.P
Soln:
Assume first term = a
Common ratio = r
S2=−4=a(1–r2)1–r ….. (1)
a5=4×a3 ⇒ar4=4ar2⇒r2=4 r=±2
From (1), we get
−4=a[1–(2)2]1–2forr=2 −4=a(1–4)−1
-4 = a(3)
a=−43
Also, −4=a[1–(−2)2]1–(−2)forr=−2 −4=a(1–4)1+2 −4=a(−3)3
a = 4
Hence, required G.P is −43,−83,−163…or4,−8,16,−32….
Q17. If p, q, and s are fourth, tenth and sixteenth terms of G.P prove that p, q, s are in G.P
Soln:
Let
First term = a
Common ratio = r
According to given condition,
a4 = ar3 = p ………. (1)
a10 = ar9 = q ………. (2)
a16 = ar15 = x ………. (3)
eqn (2) divided by eqn (1)
qp=ar9ar3⇒qp=r6
eqn (3) divided by eqn (2)
sq=ar15ar9⇒sq=r6 qp=sq
Q18. 8, 88, 888, 8888, …… are in sequence find sum of the nth terms
Soln:
Given sequence 8, 88, 888, 8888, ……
This sequence isn’t a G.P
It may be changed to a G.P if the terms are written as Sn = 8 + 88 + 888 + 8888 + ….. to n terms
=89[9+99+999+9999+……….tonterms] =89[(10–1)+(102–1)+(103–1)+(104–1)+……….tonterms] =89[(10+102+….nterms)–(1+1+1……….tonterms)] =89[10(10n–1)10–1–n] =89[10(10n–1)9–n] =8081(10n–1)–89n
Q19. If 2, 4, 8, 16, 32 and 128, 32, 8, 2, ½ are in sequences find the sum of the products
Soln:
Required sum = 2×128+4×32+8×8+16×2+32×12 =64[4+2+1+12+122]
Here, 4,2,1,12,122 is a G.P
Now first term, a = 4
And common ratio, r=12
We know that,
Sn=a(1–rn)1–r S5=4[1–(12)5]1–12=4[1–132]12=8(32–132)=314
Required sum = 64(314)=(16)(31)=496
Q20. Prove that product of corresponding terms from a, ar, ar2,… arn-1 and A, AR, AR2,… ARn-1 are in sequences find common ratio
Soln:
To show = aA, ar(AR), ar2 (AR2), … arn – 1 (ARn – 1) is a G.P series
SecondtermFirstterm=arARaA=rR ThirdtermSecondterm=ar2AR2arAR=rR
Hence, the sequences are in G.P and rR is common ratio.
Q21. If in a G.P of 4 numbers 3rd term is greater to 1st term by 9, and the 2nd term is greater to 4th by 18.
Soln:
Let
First term = a
Common ratio = r
a1 = a, a2 = ar, a3 = ar2, a4 = ar3
From given conditions,
a3 = a1 + 9 => ar2 = a + 9 ………. (1)
a2 = a4 + 18 => ar = ar3 + 18 ………. (2)
From eqn (1) and eqn (2), we get
a(r2 – 1) = 9 ……… (3)
ar(1 – r2) = 18 ……..(4)
Dividing eqn (4) by eqn (3), we get
ar(1–r2)a(1–r2)=189
=> – r = 2
=> r = -2
Substituting r = -2 in eqn (1), we get
4a = a + 9
=> 3a = 9
=> a = 3
Hence, first four terms of G.P are 3, 3(-2), 3(-2)2, and 3(-2)3
i.e. 3, -6, 12, -24
Q22. If in a G.P a, b and c are the pth, qth and rth terms. Prove that
aq – r.br – p.cp – q = 1
Soln:
Let
First term = a
Common ratio = r
From the given condition
ARp – 1 = a
ARq – 1 = b
ARr – 1 = c
aq – r.br – p.cp – q
=Aq–r×R(p–1)(q–r)×Ar–p×R(q–1)(r–p)×Ap–q×R(r–1)(p–q) =Aq–r+r–p+p–q×R(pr–pr–q+r)+(rq–r+p–pq)+(pr–p–qr–q)
= A0 x R0
= 1
Hence proved
Q23. If in a G.P ‘a’ and ‘b’ are the first and nth term and product of nth terms is P.
Show P2 = (ab)n
Soln:
Given:
First term = a
Last term = b
Therefore,
G.P = a, ar, ar2, ar3 . . . arn – 1, where r is the common ratio.
b = arn – 1…….. (1)
P = product of n terms
= (a)(ar)(ar2)…(arn – 1)
= (a x a x …. a)(r x r2 x … rn – 1)
= an r 1 + 2 + … (n – 1) …… (2)
Here, 1, 2, …..(n – 1) is an A.P
1 + 2 + …. + (n – 1)
=n–12[2+(n–1–1)×1]=n–12[2+n–2]=n(n–1)2 P=anrn(n−1)2 P=a2nrn(n−1) =[a2rn–1]n
= (ab)^{n}
Hence proved
Q24. Prove that the ratio of sum of first n terms to sum of terms from (n + 1)th to (2n)th terms is 1/ rn .
Soln:
Let
First term = a
Common ratio = r
Sum of the first n terms = a(1−rn)(1–r)
Since there are ‘n’ number of terms from (n + 1)th to (2n)th term,
Sum of the terms
Sn=an+1(1–rn)1–r
Required ratio = a(1–rn)(1–r)×(1–r)arn(1–rn)=1rn
Thus, the ratio is 1/rn
Q25. Prove that:
(a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc – cd)2
If a, b, c and d are in a G.P series
Soln:
a, b, c and d are in a G.P series
Therefore
bc = ad …………….. (1)
b2 = ac ……………… (2)
c2 = bd ……………… (3)
It is to be proven that,
(a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc – cd)2
R.H.S
= (ab + bc + cd)2
= (ab + ad + cd)2 [From (1)]
= [ab + d(a + c)]2
= a2 (b2) + 2abd(a + c) + (d2)(a + c)2
= (a2) (b2) + 2a2bd + 2acbd + (d2)(a2 + 2ac + c2)
= a2b2 + 2a2c2 + 2b2c2 + d2a2 + 2d2b2 + d2c2 [From (1) and (2)]
= a2b2 + a2c2 + a2c2 + b2c2 + b2c2 + d2a2 + d2b2 + d2b2 + d2c2
= a2b2 + a2c2 + a2d2 + b2 x b2 + b2c2 + b2d2 + c2b2 + c2 x c2 + c2d2
[From eqn (1) and (2)]
= a2 (b2 + c2 + d2) + b2 (b2 + c2 + d2) + c2 (b2 + c2 + d2)
= (a2 + b2 + c2) (b2 + c2 + d2) = L.H.S
L.H.S = R.H.S
Q26. Between 3 and 81 insert two numbers such that the sequence is a G.P series.
Soln:
Let the numbers be g1 and g2
So from the given condition 3, g1, g2, 81 is in G.P
Assume first term = a
Common ratio = r
Therefore 81 = (3)(r)3
=> r3 = 27
r = 3 (taking real roots only)
For r = 3,
Q26.
G1 = ar = (3)(3) = 9
G2 = ar2 = (3)(3)2 = 27
Hence, the numbers are 9 and 27
Q27. If a2+1+bn+1an+bn is the geometric mean for a and b. What is the value of n?
Soln:
Mean of a and b is ab−−√
From the given condition a2+1+bn+1an+bn=ab−−√
Squaring both the sides, we get
(a2+1+bn+1)2(an+bn)2=ab
=> a2n +1 + 2an + 1 bn + 1 + b2n + 1 = (ab)(a2n + 2anbn + b2n)
=> a2n +1 + 2an + 1 bn + 1 + b2n + 1 = a2n +1b + 2an + 1 bn + 1 + ab2n + 1
=> a2n +1 + b2n + 1 = a2n +1b + ab2n + 1
=> a2n +1 – a2n +1b = ab2n + 1 – b2n + 2
=> a2n +1 (a – b) = b2n + 1 (a – b)
=> (ab)2n+1=1=(ab)0
=> 2n + 1 = 0
=> n=−12
Q28. Prove that the ratio of two numbers is (3+22–√)(3–22–√) if the sum of the two numbers is 6 times their geometric mean.
Soln:
Let the two numbers be a and b.
G.M = ab−−√
From given conditions,
a+b=6ab−−√ ……. (1)
=> (a + b)2 = 36(ab)
Also,
(a – b)2 = (a + b)2 – 4ab = 36ab – 4ab = 32 ab
=> a–b=32−−√ab−−√
= 42–√ab−−√ ……. (2)
Eqn(1) + Eqn (2)
2a=(6+42–√)ab−−√
=> a=(3+22–√)ab−−√
Putting the value of a in eqn (1),we get
b=6ab−−√–(3+22–√)ab−−√
=> b=(3–22–√)ab−−√ ab=(3+22√)ab√(3–22√)ab√=3+22√3–22√
Hence, required ratio = (3+22–√):(3–22–√)
Q29. For two positive numbers if G and A are the G.M and A.M show that the numbers are A±(A+G)(A–G)−−−−−−−−−−−−√
Soln:
Given G and A are A.M and G.M
Assume the numbers be a and b.
AM=A=a+b2………(1) GM=G=ab−−√………(2)
From Eqn(1) and Eqn (2), we get
a + b = 2A ……. (3)
ab = G2 ……. (4)
Putting the value of a and b from (3) and (4)
(a – b)2 = (a + b)2 – 4ab, we get
(a – b)2 = 4A2 – 4G2 = 4(A2 – G2)
(a–b)=2(A+G)(A–G)−−−−−−−−−−−−√……(5)
From eqn (3) and (5), we get
2a=2A+2(A+G)(A–G)−−−−−−−−−−−−√
=> a=A+(A+G)(A–G)−−−−−−−−−−−−√
Putting the value of a in eqn(3), we get
b=2A–A–(A+G)(A–G)−−−−−−−−−−−−√=A–(A+G)(A–G)−−−−−−−−−−−−√
Hence, the numbers are A±(A+G)(A–G)−−−−−−−−−−−−√
Q30. Bacteria over certain culture multiply twice in one hour. If in initial stage there were only 30 bacteria, what will be the amount of bacteria in the 2nd, 4th and nth hour?
Soln:
Given that amount of bacteria multiplies two times in one hour
Therefore,
Bacteria amount will form G.P
Here, first term = (a) = 30
Common ratio = (r) = 2
Therefore ar2 = a3 = 30 x (2)2 = 120
Therefore amount of bacteria after 2nd hour is 120
ar4 = a5 = 30 x (2)4 = 480
Therefore amount of bacteria after 4th hour is 480
arn x an + 1 = 30 x (2)n
Hence, amount of bacteria after nth hour is 30 x (2)n
Exercise 9.4
Q1. Find sum to nth term for the series 1×2+2×3+3×4+4×5+....
Soln:
Given series = 1×2+2×3+3×4+4×5+.... nth term, an = n (n + 1)
Sn=∑nk=1a1=∑nk=1k(k+1) =∑nk=1k2+∑nk=1k n(n+1)(2n+1)6+n(n+1)2 n(n+1)2(2n+13+1) n(n+1)2(2n+43) n(n+1)(n+2)3
Q2. Find sum to nth term of the series 1×2×3+2×3×4+3×4×5+.....
Soln:
Given series = 1×2×3+2×3×4+3×4×5+...nthterm,an=n(n+1)(n+2)
= (n2 + n) (n + 2)
= n3 + 3n2 + 2n
S0=∑nk=1ak =∑nk=1k3+3∑nk=1k2+2∑nk=1k =[n(n+1)2]2+3n(n+1)(2n+1)6+2n(n+1)2 =[n(n+1)2]2+n(n+1)(2n+1)2+n(n+1) =n(n+1)2[n(n+1)2+2n+1+2] =n(n+1)2[n2+n+4n+62] =n(n+1)4(n2+5n+6) =n(n+1)4(n2+2n+3n+6) =n(n+1)[n(n+2)+3(n+2)]4 =n(n+1)(n+2)(n+3)4
Q3. Find sum to nth term of the series 3×12+5×22+7×32+...
Soln:
Given series = 3×12+5×22+7×32+...nthterm,
an = (2n + 1) n2 = 2n3 + n2
Sn=∑nk=1ak =∑nk=1=(2k3+k2)=2∑nk=1k3+∑nk=1k2 =2[n(n+1)2]2+n(n+1)(2n+1)6 =n2(n+1)22+n(n+1)(2n+1)6 =n(n+1)2[n(n+1)+2n+13] =n(n+1)2[3n2+3n+2n+13] =n(n+1)(3n2+5n+1)6
Q4. Find sum to nth term of the series 11×2+12×3+13×4+....
Soln:
Given series = 11×2+12×3+13×4+.... nthterm,an=1n(n+1)=1n–1n+1 a1=11–12 a2=12–13 a3=13–14... an=1n–1n+1
Column wise adding the above terms, we get
a1+a2+a3+…+an=[11+12+13+...+1n]–[12+13+14+...+1n+1+] Sn=1–1n+1=n+1–1n+1=nn+1
Q5. Find sum to nth term of the series 52+62+72+...+202nthterm,an=(n+4)2=n2+8n+16
Soln:
Sn=∑nk=1ak=∑nk=1(k2+8k+16) Sn=∑nk=1ak=8∑nk=1k+∑nk=116 =n(n+1)(2n+1)6+8n(n+1)2+16n
16th term is (16 + 4)2 = 202
S16=16(16+1)(2×16+1)6+8×16×(16+1)2+16×16 =(16)(17)(33)6+(8)×16×(16+1)2+16×16 =(16)(17)(33)6+(8)(16)(17)2+256
= 1496 + 1088 + 256
= 2840
52 + 62 + 72 + …. + 202 = 2840
Q6. Find sum to nth term of the series 3×8+6×11+9×14+…
Soln:
Given series = 3×8+6×11+9×14+…an
= (nth term of 3, 6, 9, …) x (nth term of 8, 11, 14 .. .)
= (3n) (3n + 5)
= 9n2 + 15n
Sn=∑nk=1ak=∑nk=1(9k2+15k) =∑nk=1k2+15∑nk=1k 9×n(n+1)(2n+1)6+15×n(n+1)2 =3n(n+1)(2n+1)2+15n(n+1)2 =3n(n+1)2(2n+1+5) =3n(n+1)2(2n+6) =3n(n+1)(n+3)
Q7. Find sum to nth term of the series 12+(12+22)+(12+22+32)+....
Soln:
Given series = 12+(12+22)+(12+22+32)+....an =(12+22+32+....+n2) =n(n+1)(2n+1)6 =n(2n2+3n+1)6 =2n3+3n2+n)6 =13n3+12n2+16n Sn=∑nk=1ak =∑nk=1(13k3+12k2+16k) =13∑nk=1k3+12∑nk=1k2+16∑nk=1k =13n2(n+1)2(2)2+12×n(n+1)(2n+1)6+16×n(n+1)2 =n(n+1)6[n(n+1)2+2n+12+12] =n(n+1)6[n2+n+2n+1+12] =n(n+1)6[n2+n+2n+22] =n(n+1)6[n(n+1)+2(n+1)2] =n(n+1)6[(n+1)(n+2)2] =n(n+1)2(n+2)12
Q8. Find sum to nth term of the series that has nth term given by n(n + 1)(n + 4)
Soln:
an=n(n+1)(n+4)=n(n2+5n+4)=n3+5n2+4n Sn=∑nk=1ak=∑nk=1k3+5∑nk=1k2+4∑nk=1k =n2(n+1)24+5n(n+1)(2n+1)6+4n(n+1)2 =n(n+1)2[n(n+1)2+5(2n+1)6+4] =n(n+1)2[3n2+3n+20n+10+246] =n(n+1)2[3n2+23n+346] n(n+1)(3n2+23n+34)12
Q9. Find sum to nth term of the series that has nth term given by n2 + 2n
Soln:
an = n2 + 2n
Sn=∑nk=1k2+2k=∑nk=1k2+∑nk=12k ……………(1)
Take ∑nk=12k=21+22+22+....
The series 2, 22, 23, …. Is a G.P with 2 as first term as well as common ratio.
∑nk=12k=(2)[(2)n–1]2–1=2(2n–1) ……………… (2)
From (1) and (2), we get
Sn=∑nk=1k2+2(2n–1)=n(n+1)(2n+1)6+2(2n–1)
Q10. Find sum to nth term of the series that has nth term given by (2n – 1)2
Soln:
an = (2n – 1)2 = 4n2 – 4n + 1
Sn=4∑nk=1ak–4∑nk=1(4k2–4k+1) =4∑nk=1k2–4∑nk=1k+∑nk=11 =4n(n+1)(2n+1)6–4n(n+1)2+n =2n(n+1)(2n+1)3–2n(n+1)+n =n[2(2n2+3n+1)3–2(n+1)+1] =n[4n2+6n+2–6n–6+33] =n[4n2–13] =n(2n+1)(2n–1)3
0 Comments