Class 11 Chapter 9_ Sequences and Series (Geometric Progressions) (sum of n terms)_ Lecture 7.pdf

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Class 11 Chapter 9 Sequences and Series_ Lecture 7
#geometric_progression
#sum_of_finite_terms
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1. Dr. Pranav Sharma
Maths Learning Centre. Jalandhar.

2. nth TERM FROM THE END OF A GP
Show that the uth term from the end of a GP with the first term a, the common ratio

re, |A U
r and the last term | is given by 3.

Let a be the first term, r be the common ratio and I be the Last term of a given GP.
l

L 1 L
Then, 2nd term from the end == aay 3rd term fromthe end = EG

1
..nth term from the end = Fa)

1
Hence, the nth term from the end =p
r

Find the gth term from the end of the GP3, 6,12, 24,... 12288.
Here r = 2 and l = 12288.

_ 12288 12288 —

th E albania le,
8" term from the end = aay = 7 = 57 = 28 — 96.

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3. PROBLEMS ON finding the GP

For solving problems on GP, it is always convenient to take:

(t) 3 numbers in GP as a a, ar;

it) 4 numbers tw GPs war, FU

(
(ii) 5 numbers in GP as + 2 a ar, ar’;
(

v) the terms as a, ar, ar”, oi their product is not given.

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4. 13 > . >
The sum of first three terms of a GP ist and their product is — 1. Find these terms.

Let the first three terms of the given GP be a a and ar.

Temixaxar==1>a=-1>a=-=1.
a 13 -1 13
A Sehr = Lor = [a==1]
rire ac Aa 25r = —12 = 12r?
Tr 7712 a ls il

= 121? +25r +12 = 0 = 121? + 16r+9+12=0
> 4r(3r + 4) + 3(3r + 4)

> (3r +4)(4r +3) =0=7r=
So, the required terms are

fs fes 19 te È-1

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5. Find three numbers in GP. deta sum is13 Ada the sum of whose squares is 91.

Let we required numbers bei =" a and ar. Then, © +a+ar=13 (i)

and“ = Eh a? + ar? = 91..... (ti) On squaring both sides of (i), we get

a 2 a? 53 a?
Eat) = 169 > (Gat sar?) 2 (E +a +a *r) = 169
>91+ 2a (2 + a+ ar) = 169 [using (4)1 > 91 + 26a = 169 [using (i)]
> 26a=78>a=3.

Putting a = 3 in (1), we get
3 3
7+3+3r=13>7+3r=10>3r*-10r+3=0

2(-36r-D=0>r=301=5
Hence, the required numbers are 1,3, 9 or 9,3, 1.

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6. Find three numbers in GP. whose sum is 52 and the sum of whose products in pairs
is 624.
Let the required numbers be a, ar, ar”. Then,
(a+ ar + ar”) =52 =a(1+r+r?) =52 (t)

and a-ar+ar-ar?+a-ar? =624 > aAr(14+r+r?)=624. (it)
On dividing (ii) by (t), we get ar = 12 and therefore, a = =
Putting a = in (0), we get = (1+r+12)=52>23(1+r+r?)=13r

= 37? -101+3=0>(8r-D(173)=0>r=jor1=3.
a=360a=4.
Hence, the required numbers are 36, 12, 4 or 4,12, 36.

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7. if the product of three numbers in GP is 216 and the sum of their products in pairs
is 156, find the numbers.

Let the required numbers vel, a and ar. Then,
a
7X ax ar = 216 = a =216= 69 >a=6.
And Ex a+ a X ar +%x ar = 156
1
Sa (G+r+1) = 156 > (67)(1+7r+7r?) = 156r [a = 6]
> 36(12+r+1)=156r > 3(r? +741) = 13r
>31?-10r+3=0>(8r-D(r-3)=0>r=Zorr=3.
So, the required numbers are 18, 6,2 or 2, 6, 18.

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8. Find four numbers in GP such that the third term is greater than the first by 9 and

the second term is greater than the fourth by 18.

Let the required numbers be a, ar, ar? and ar?. Then,
T3-T,=9>ar?-a=9>a(r?-1)=9 ()

and T¿ —T¿=18 > ar — ar? = 18 = ar(1—r?) = 18. (i)

On dividing (ii) by (1), we get r = —2.

Putting r = —2 in (i), we get a(4 — 1) =9 > a=3.

Hence, the required numbers are 3, —6,12 and —24.

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9. The sum of three numbers in GP. is 56. If we subtract, 7, 21 from these numbers
in that order, we get an AP. Find the numbers.
Let the required numbers be a, ar and ar?. Then, a + ar + ar? = 56. (i)
Also, (a — 1), (ar — 7) and (ar? — 21) are in AP.

So, 2 (ar —7) = (a—1) + (ar? — 21) > a+ ar? = 2ar +8. (i)
using (it) tw (I), we get ar + (Zar + 8) = 56 > 3ar=48 > ar=165r= =

Putting r = in (0), we get a + 16 + 2° = 56 = a? + 16a + 256 = 56a
> a? - 40a + 256 = 0 = a? - 8a — 32a +256 = 0
= a(a— 8) —32(a-8) = 0 => (a- 8)(a— 32) =0

> a=8ora= 32.

16 16 1
Now, 458231 ==3>1=2. And, a=32 231 =2T1=35

7
Hence, the required numbers are (8, 16, 32) or (32, 16, 8).

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16

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10. SUM OFN TERMS OF A GP
Prove that the sum of n terms of a GP with the first term a and the common ratior
na, whenr = 1;
a(i-r")
is given by Sn = À An
a(r"-1)
ED,
Let us consider a GP with the first term a and the common ratio r. Then,
Sh =atar+ar?+--+ar™1, (i)
Casettfr = 1 we have S, = a+ a + ton terms = na.
case2 fr # 1, we have rS, = ar + ar? ++ ar"! + ar, (it)
On subtracting (tt) from (1), we get (1 — r)S, = (a — ar”) = a(1-—r”)
a(i-r” a(ri-1
Sn = = or Sn = a

, whenr < 1;

whenr > 1.

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11. tf a GP contains n terms with first term = a, common ratio = r and last term = I
then l=ar"™!, (i)

Case 1 When r < 1, we have
a(i-r") _ (a-ar") _ (a-Ir)

Sa an "an an Ming OI.
Case 2 when r > 1, we have

_ a(r"-1) _ (ar"-a) _ (Ir-a) : À
SGD an) > Gd sing WI.

Find the sum of 8 terms of theGP 3,6,12, 24, ...
Hee a =3,r=2>1andn=8.

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e (r"—-1)
Using the formula, Sp = =; , We get
sei. à (256=1) =3x255=765
= —_ = 3 Aa =3x = \
a 2-2
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12. 7 ndbay, 1, Mort
Find the sum of the geometric series 1 +> +7 + gr 012 terms.

Herea = Lr =5<1andn = 12.
Latin)

using the formula, S = Cony + We get
n
2 _1xf1-() } (1-31) 2-1) 4095
1g Ce = 5 im .
ED D 7 2

How many terms of the geometric series 1+4+16+64+ - will make the sum 5461?
Let the required number of terms ben. Here, a=1,r=4> land S, = 5461.
5, = DL AD ggg (47 — 1) = 16383
™™ (r-1) (4-1) u
> 4" = 16384 =47> n=7.
Hence, the required number of terms is 7.

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13. Find the sum of the series 2 +6 + 18 + 54 +1" +4374.
Clearly, the given series is a geometric series tw which a = 2,r=3>1andl=
4374.

4 o (ira) (4374x3-2) _ 13120
the required sum = eo om 2
Hence, the sum of the given series is 6560.

= 6560.

In a GP, itis being given thatT, = 3, Tn = 96 and Sn = 189. Find the value of n.

Here, a = 3,1 = 96 and Sn = 189.

Let the common ratio of the given GP ber. Then, Sy = “ne —_ = 189

> (96r — 3) = (189r — 189) > 93r = 186 > r =2.
Now,I= ar 1>3x 2271 = 96 = 27-1 = 32 = 28 sn-1=5>n=6.
Hence, n = 6.

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14. Sum the series 5+55+555+ + ton terms.
We have 5+55+555+... ton terms = 5 X{1 + 11 + 111 +... ton terms}

= Ex + 99 + 999 + to n terms)

= = x{(10 — 1) + (10? — 1) + (103 — 1) + ton terms)
= 2x {(10 + 10? + 10° + to nterms)—n}

_ 5 (10x(10"-1) 5
=: = = n} == x (10"* - 9n — 10).

Hence, the required sum is = x (10"*! — On — 10).

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15. Sum the series 0.4+ 0.44 + 0.4444... ton terms.
we have 0.4 + 0.44 + 0. 444+...to n terms

=4x( 14.114.111 + ton terms}= 5 x{.9+.99+.999 + --- tom termas)
= $x{(1-.1) + (1-.01) + (1.001) + ton terms)

= 2 X{ (1+1-+tonterms)-(1+.01+.001+... to n terms)}

4 1x{1-.(2} aG-r")) 4 aa
-§|n- a=D | (es An } = 5x contre dae]
5-3)
=a Pn fi] - “a x [on - 1+ 55}

Hence, the required sum isz Pres (on- le

107

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16. The sum of some terms of AGP is 315. its first termis 5 andthe common ratio is 2.
Find the number of its terms and the last term.
Let the given GP contain n terms. Then,
a=5,r=2>1andS, = 315.

a(r" — 1) 5 x (27 —1)

S, = 315 =D = 315 2-1 = 315
Deeg Er 22 ES Se!

Last term = ar®-D = 5 x 2670 = (5 x 25) = (5 x 32) = 160.
Hence, the given GP contains 6 terms and its Last term is 160.

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17. The sum of first three terms ofa GP ís10 and the sum of its next three terms is 128.
Find the sum ofn terms of the GP.

Let a be the first term and r be the common ratio of the given GP.

Then, a+ ar + ar? = 16 and ar? + art + ar? = 128

=a(l+r+r?)=16() and ar$(1+r+72) = 128 (ii).

on dividing (ti) by (i), we get Bal? Dn

Putting r = 2 in (i), we get

16
a(l+2+4)=16>a=—.

Thus, in the given GP, we have a = = andr =2>1.

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16
0) PAD _ 16 90 _
Sua ED QQ 7 Arrmäte
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18. ma GP, the sum of first two terms is—4 and the sth term is 4 times the srd term.
Find the GP.

Let a be the first term and r be the common ratio of the given GP. Then,
T¡+T2=-4mmdT5¿=4xT3 >a+ar=-4andart=4xar?

aal+r)=-4madr?=4 >a(1+r)=-4 (i) andr = +2.
|

Putting T.= 2 in (I), we get a=
Putting r = —2 in (i), we get a = 4.

a= = and r = 2 gives the ez, 3%, _
and, a = 4 and r = —2 gives the GP 4, —8, 16, ...
-4 -8 -16

Hence, the required GP is { } or {4,-8,16,...}.

CUE

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19. in an increasing GP, the sum of the first and last terms is 66, the product of the
second and the last but one is128 and the sum of the terms is 126. How many
terms are there in this GP?
Let the given GP contain n terms. Let a be the first term and r be the common ratio
of this GP. Since the given GP is increasing, we have r > 1.
Now, T1 + T, = 66>a+ ar™-) = 66, (i) And,
TT, 28 a0 zo? = 1202 wr? = 128008 = (it)
using (ti) tn (D, we get a + À = 66 = a? — 66a + 128 = 0
>a? — 2a — 64a + 128 = 0 = a(a— 2) — 64(a— 2) =0
> (a-2)(a-64)=0>a=2ora= 64.
, Te (n-1) _ 128 _ 128 _ _
Putting a = 2 in (it), we getr' = 7 = Watt 32 a rmiCl4L)
mn 220, et

Ea az which is reiected, since

Putting a = 64 in (it), we get r
r> 1.
Thus, a = 2 and r@D = 32,

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20. lA arm) Cas AN rd C2
Now, 5, = 12635212632 (7) = 126 5 St 63
xr 1_ a 82r-1
r-1 u r-1..

>32r-1=-63r-63>31r=62>r=2.
pod. 92 5h 1= Fh 6,
Hence, there are 6 terms in the given GP.

Consider two GPs a, ar, ar?, ar?,... and A, AR, AR?, AR?, ...
Taking the products of their corresponding elements, we get

aA, aA(rR), aA(rR)?, aA(rRY, ...
which is clearly a GP with first term = aA and common ratio= rR.

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21. Find the sum of the products of the corresponding terms of finite geometrical
progressions 2, 4,8, 16,32 and 128, 32,8, 2, 5

Taking the products of the corresponding terms of two given GPS, we get a new
progression (2 x 128), (4 x 32), (8 x 8), (16 x 2), (32 x),

L.e., 256, 128, 64, 32, 16, which is clearly a GP in which a = 256 and
16 1

ra gel

ars) _ 256%{1-(3) | _ 250x(1-3) 1
the required sum = "Gay = 13) = 0 =256x2.x (1 > =)

2.

(256 x2 >) 496
= sao
32

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22. Find the sum of n terms of the sequence given by an = (3" + 5n), n EN,
Let the sum of n terms of the given sequence be Sn. Then,

Sy = 41 +a, +a3+--+a,
=(31+5x1)+(32+5x2)+(3+5x3)+-+(37+5x0m)
=(34+324+334+-43)+(5x1+5x2+5x3+--+5xm)

= (3432433 +43) 4+5x(142+3+-:+m)
3(3" — 1)

„8-9

=3(3"-1) +3xnm+1),

+5x7(1+m)

Hence, the required sum is : (3"-1)+ En(n +1).

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23. (f S13) S2 and Sz be respectively the sum of n, Zn and 3n terms of a GP. then prove

<
that S1(S3 — Sz) = (S2 - $1)? 9
Let a be the first term and r be the common ratio of the given GP. Then, Es
dE Tr aa”), [ae oO
S1(S3 — 82) = un” { (1=r) Ar) } 8
_a(í-r") | (a-ar?"-a+tar?*) _ a(i-r) | ari) _ a2r2(1-r")2 3
== (Ar) an) an). 1 A2 ®
612 _ fatter") ary)? _ (a-ar2"-a+ar") array zZ
And, (Sz = 81) = { (1=r) (1=r) } = (1-r)2 (or? >
Anny ES
don? ° 3
Hence, Sy (S3 — S2) = (Sz - 51? =
5
jo)
=
5
o
El
®
2
€
=
oO
n
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24. IfS be the sum, P be the product and R be the sum of the reciprocals of n terms in a

GP, prove that P? = (Ey.

a(i—r")

Let the given GP be a, ar,ar?,ar®-), then, = ayn

(0)

and P = [ax ar x ar? x ..xar iD]

1-1 1n-1
=> P = qty l1t2+34-4+(n-1)] = aaa > P= ara Jn

1
= P2 = arn) and, R= [+44 +} = (2): bz)

aon) ~ Na) E

R= (5) TS Ra Oy + Mi) On diiding © by Ci), we get
yn a O)
S_aG@-r") aG-n:r = at)
Rie br) (1-18)

(E) =(a2r0-D)" = ar. 00-08 = p? [using (01. Hence, P2 = (£)"

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25. Aman has 2 parents, 4 grandparents, 8 great-grandparents, and so on. Find the
number of his ancestors during the ten generations preceding his own.
Required number of ancestors = 2+4+6+8 +... up to 10 terms

10_
=2x G2) (2104) IGP with a=2,r=2 and n = 101

(2-1)
= 2(1024 — 1) = (2 x 1023) = 2046.
Hence, the required number of the man's ancestors is 2046.

A man writes à letter to four of his friends. He asks each one of them to copy the
letter and mail to four different persons with the instruction that they move the
chain similarly. Assuming that the chain is not broken and it costs Rs4 to mail
one letter, find the amount spent on postage when eth set of letters is mailed.
Successive number of Letters are 4, 16, 64,... This is a GP with a = 4andr = 4.
Number of letters in the eth set = ar) = (4 x 45) = 46 = 4096.

Cost of postage on these Letters = Rs(4096 x 4) = Rs16384.

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26. What will RS10000 amount to in 4 years after its deposit in a bank which pays
annual interest at the rate of 10% per annum compounded annually?

10
Amount after 1 year = Rs 10000 + (10000 x =" x 1) = Rs 11000,
amount after 2 years = Rs 11000 + (11000 x =~ x 1) = Rs 12100,

amount after s years = Rs 12100 + (12100 x + x 1) = Rs 13310, and so on.

Thus, these amounts form a GP 10000, 11000, 12100, 13310,

11000 12100 13310 _ 11 =
Googe = Glue 20° a In this GP, we have a = 10000, r =

such that

Amount after 4 years = Ts = arS-) = art = Rs [10000 x (4 x e Rs14641.

Alternative method Here, P = Rs10000, R=10% p.a. and T = Fa years.
=Rs|Px(1 zy = Rs}10000 x (1 +4 at
amount after + years = Rs aoa) {= os +

4
= Rs{10000 x E) } = Rs 14641.

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27. A manufacturer reckons that the value of a machine which costs him RS125000 will
depreciate each year by 20%. Find value of the machine at the end of 5 years.

5
Value of the machine after 5 years = Rs {125000 x (1 a =) }

5
= Rs {125000 x (2) } = Rs (LAA) = Rs 40960.

Alternative method Initial value of the machine = Rs 125000.
value of the machine after 1 year = Rs (125000 x =~) = Rs 100000.

Value of the machine after 2 years = Rs (100000 x =~.) = Rs 80000.

Thus, the year wise values of the machine are Rs125000, Rs100000, Rs80000, ...

100000 _ 4 80000 _ 4 , u
2000 — 8 100006 — et Thus, these amounts form a GP with a = 125000

and r = £ Value after 5 years = 6th term = Té = ar) = ar? = Rs {125000 x

9 = Rs (SPL) = Rs 40960.

Here,

5 3125

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28. The number of bacteria in a certain culture doubles every hour. If there were 30
bacteria present in the culture originally, how many bacteria would be present at the
end of 2nd hour, 4th hour and uth hour

The bacteria present iw the culture originally, at the end of ist hour, at the end of

2nd hour, at the end of zrd hour and so on, are 30,60, 120,240, ...

This is a GP with a = 30 andr = = 2,

Number of bacteria at the end of 2nd hour= T3 = ar? = (30 x 22) = 120.
Number of bacteria at the end of 4th hour= Ts = art = (30 x 24) = 480.
Number of bactería atthe end of uth hour= Tye = art) = 30 x 27.

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29. The inventor of the chessboard suggested a reward of one grain of wheat for the first
square 2 grains for the second; 4 grains for the third, and so on, doubling the
number of grains for subsequent squares, How many grains would have to be given
to the inventor
Required number of grains = 1+ 21+ 2? + 23 +--- to 64 terms

= 14 (21427423 +.-.26)

2(263-1
=1+ = (1+ 2% = 2) = (21).

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30. The lengths of three unequal edges of a rectangular solid block are in GP. The
volume of the block is 216 cm? and the total surface area is 252 cm?. Find the
length of its longest edge.

Let the Lengths of its edges be e cm, 4 cm and Ar cm.

Then, its volume = € xax ar) cm? = a?cm3, So, a? = 216 —6? = a = 6.

So, the edges are 6r cm, 6 cm and À om.
surface area = 2[lb + bh + Ih] = 2 [6r x 6.4.6 x £ + 6r x Ÿ] cm?
36 , 1
= 2 |36r ++ 36] om? = 72 x [r+2+ 1] cm’.
1
72x(r+241)2 2523 2 (r2 + MT) HP BaF or ee BO

is

>(r-2)Qr-1)=0>r=2or=;

Each value of r gives the longest edge = 12 om.

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