Download IIT JEE sequence and series and more Study notes Mathematics in PDF only on Docsity!
1. Sequence & Series
Sequence & Series
Introduction:
A sequence is a set of terms which may be algebraic, real or complex
numbers, written according to definite rule and the series thus formed
is called a progression.
E.g.
0, 7, 26, …(rule is n 3 -1)
1, 4, 7, 10, …
2, 4, 6, 8, … etc.
Note: Minimum number of terms in a sequence should be 3.
Arithmetic Progression:
It is a sequence whose terms increase or decrease by a fixed number. Fixed
number is called the common difference. If ‘a’ is the first term and ‘d’ is the
common difference, then the standard appearance of an A. P. is
a, ( a + d) ( , a + 2 d ) , ..., ( a + n − 1 d)
n th
term = Tn = a + ( n −1 d)
Note:
If d > 0 ⇒ increasing A. P. If d < 0 ⇒ decreasing A. P.
If d = 0 ⇒ all terms remain same
Q1. If 6 th and 11 th term of an A.P. are respectively 17 and 32. Find the 20 th
term.
Sol: T 6 = 17 ⇒ a+5d = 17 …(1)
T 11 = 32 ⇒ a+10d = 32 …(2) (2) – (1)
5d = 15 ⇒ d = 3 Putting in (1)
a + 5(3) = 17 ⇒ a = 2
now, T 20 = a + 19d =2 + 19(3) = 59
Q2. In an A. P. if tp = q and tq = p, then find the r th term.
Sol: t p = q ⇒ a + (p - 1)d = q …(1)
tq = p ⇒ a + (q - 1)d = p …(2)
(1) – (2) q – p = (p – q)d
d = –
Sequence & Series 2.
from (1), a + (p - 1)(-1) = q ⇒ a = p + q - 1 Now
tr = a + (r – 1)d = (p + q - 1) + (r - 1)(-1) = p + q -r
Q3. In an A. P. if a 2 + a 5 – a 3 = 10 and a 2 +a 9 = 17, then find the 1 st term and
the common difference Sol: a 2
(a + d) + (a + 4d) - (a + 2d) = 10 a + 3d = 10 …(1)
a 2 + a 9 = 17
(a + d) + (a + 8d) = 17 2a + 9d = 17 …(2)
(2) – 2 × (1) 3d = -3 ⇒ d = -1, a = 10 - 3d = 13
Q4. If p th , q th & r th term of AP are respectively a, b and c, then prove that
a q ( − r) + b r( − p ) + c p( − q) = 0
Sol: T p ⇒ A + (p - 1) D = a …(1)
Tq ⇒ A + (q - 1) D = b …(2) T r ⇒ A + (r - 1) D = c …(3)
(1) - (2) ⇒ (p - q) D = a - b
(2) - (3) ⇒ (q - r) D = b - c (3) - (1) ⇒ (r – p)D = c – a
a(q-r)+b(r-p)+c(p-q)
a b ( c) b c( a) c a( b)
D D D
= ab-ac+bc-ab+ac-bc D
= 0 H.P.
Q5. If a, b, c, d are distinct integers in an A.P. such that d = a^2 + b^2 + c^2 ,
then find the value of a + b + c + d.
Sol: Let b = a + α, c = a + 2α, d = a + 3α now d=a^2 + b^2 + c^2
a + 3α = a 2
a + 3α = 3a^2 + 6aα + 5α^2
⇒ 5 α 2
- 3(2a – 1)α + a(3a – 1) = 0 …(1)
now α ∈ R ⇒ D ≥ 0 ⇒ 9(2a - 1)^2 - 4.5.a(3a - 1) ≥ 0
⇒ 36a 2
- 36a + 9 - 60a 2 + 20a ≥ 0 ⇒ 24a^2 + 16a – 9 ≤ 0
a , 12 12
∵ a ∈ I hence a = 0 or -
Sequence & Series 4.
Q2. The sum of first 3 terms of an AP is 27 and the sum of their squares is 293 then find S n Sol: a + (a + d) + (a + 2d) = 27 ⇒ 3(a + d) = 27 ⇒ a + d = 9 …(1)
now, a 2
- (a + d) 2
- (a + 2d) 2 = 293
(9-d) 2
- 9 2
- (9+d) 2 = 293 243 + 2d^2 = 293 ⇒ 2d^2 =
d 2 = 25 ⇒ d = ± 5 from (1)
if d = 5 ⇒ a = 4
if d = -5 ⇒ a = 14
Now, for a = 4, d = 5 ⇒ n ( ) ( )
n n S 8 n 1 5 3 5n 2 2
For n ( ) ( ) ( )
n n a 14, d 5 S 28 n 1 5 33 5n 2 2
Q3. In an AP of which ‘a’ is the 1st^ term, if the sum of the 1st^ p terms is equal
to zero, show that the sum of the next q terms is
aq p( q)
p 1
Sol: Given Sp = 0
p 2a p 1 d 0 2a p 1 d 0 2
2a d p 1
now sum of next q terms
=Tp+1+ Tp+2+…(q terms)
q 2 a pd q 1 d 2
q 2a 2a 2p q 1 2 p 1
^ −
q 2ap 2a 4ap 2aq 2a
2 p 1
q 2a p ( q) p q
aq 2 p 1 p 1
−^ −
5. Sequence & Series
Q4. Find the value of the sum
359
k 0
k cosk
∑ °
Sol: Sum
S = 0cos0 ° + 1cos1 ° + 2cos2 ° + 3cos3 ° + ... + 359cos359°
∵cos359° = cos1°
cos358 cos
cos181 cos
hence
S = 360 cos1 { ° + cos2 ° + … + cos179°} + ( cos180°) 180
n=179, α = 1, β = 1
sin 2 S 360 cos 90 180 1 sin 2
= × ° −
S = - 180
Q5. In an AP if Tm =1/n and Tn = 1/m then show that mn ( )
S mn 1 2
Sol: m ( )
T a m 1 d n
n^ (^ )
T a n 1 d m
m n 1 m n d d mn mn
from (1)
a m 1 a mn n n mn n
a mn
now (^) mn { ( ) }
mn S 2a mn 1 d 2
mn 2 mn 1 1 mn 1 2 mn mn 2
7. Sequence & Series
Q8. If S ,S ,S , … ,S 1 2 3 p are the sum of n terms of ‘p’ arithmetic series whose
first terms are 1,2,3,4,… and whose common differences are 1,3,5,7,…
prove that
1 2 3 p^ (^ )
np S + S + S +…+S = np 1 2
Sol: Clearly
1 {^ (^ ) }
n S 2 1 n 1. 2
= × + −
2 {^ (^ ) }
n S 2 2 n 1. 2
= × + −
3 {^ (^ ) }
n S 2 3 n 1. 2
= × + −
p {^ (^ ) (^ )}
n S 2 p n 1. 2p 1 2
+ = × + − −
1 2 p {^ (^ )^ (^ )^ (^ (^ ))}
n S S ...S 2. 1 2 3 ... p n 1 1 3 5 ... 2p 1 2
n p p^12
- n 1 p 2 2
np np 1 H.P. 2
Q9. In an AP S p = q and S q = p then show that S p+q = - (p + q)
Sol: S p = q
⇒ (^) { ( ) }
p 2a p 1 d q 2
d ap p p 1 q 2
also S q = p
⇒ (^) { ( ) }
q 2a q 1 d p 2
d aq q q 1 p 2
⇒ a(p - q) + {(p 2
d
2
= - (p - q)
⇒ a+(p+q-1)
d
2
Sequence & Series 8.
⇒ 2a+(p+q-1)d = - 2 …(3)
now
p q {^ (^ ) }
p q S 2a p q 1 d 2
p q 2 p q 2
= − = − + H.P.
Q10. Let a , a , a , ..., a 1 2 3 100 be an arithmetric progression with a 1 = 1 and
p
p i i 1
S a
= (^) ∑ , 1 ≤ p ≤ 100. For any integer n with 1 ≤ n ≤ 20 , let m = 5n. If
m
n
S
S
does not depend on n then a 2 is.
Sol:
{ (^ ) }
{ ( ) }
1 5n^1
n 1 1
5n 2a 5n 1 d S 2a^ d^ 5nd 2 5 S n 2a d nd 2a n 1 d 2
−^ +
It is independent of n if
(i) d = 0 or (ii) 2a 1 = d ⇒ d = 2
if d = 0 then a 2 = a 1 = 1
if d = 2 then a 2 = a 1 +d = 3
Q11. The number of terms in an AP is even, the sum of the odd terms is 24,
sum of the even terms is 30, and the last term exceeds the first term
by
. Find the number of terms.
Sol: let AP is a , a , a , ..., a 1 2 3 2n
sum of odd terms
a 1 + a 3 + a 5 + ... + a2n (^) − 1 = 24 …(1)
sum of even terms
a 2 + a 4 + a 6 + ... + a2n = 30 …(2)
( a 2 −^ a 1 ) +^ ( a^4 −^ a 3 ) +^ ...^ +^ ( a2n^ −^ a2n^ − 1 )=^6
nd = 6 … (3)
now given
2n 1
a a 10 2 2
2n 1 d 2
Sequence & Series 10.
( (^ ) ) (^ (^ ))
2 2 = 5 n − n − 1 + 2 n − n − 1
= 5 2n ( − 1 ) + 2 = 10n − 3;n ≥ 2
hence sequence is
11, 17, 27, 37, …, (where T 1
= S
1
Q14. Evaluate:
S= 2
- 99 2 + 98 2 - 97 2 + 96 2 - 95 2 +…+ 2 - 1 2
Sol: S = (100-99)(100+99) + (98-97)(98+97) +…+ (2-1)(2+1)
S = 100+99+98+97+…+2+
= × =
Q15. Suppose a , a , ... 1 2 are in A.P and Sk denotes the sum of the first k
terms of this AP. If
2 n 2 m
S n
S m
= for all m, n ∈ N, then prove that
m 1
n 1
a 2m^1
a 2n 1
Sol:
2 n^1 2 m (^1)
S n / 2^ 2a^ n^ 1 d n
S m / 2 2a m 1 d m
1
1
2a n 1 d (^) n
2a m 1 d m
1
1
n 1 a d 2 n
m (^1) m a d 2
In (1) replace n→ 2n+
m→ 2m+
⇒
( )
( )
1
1
a nd 2n^1
a md 2m 1
Hence,
m 1
n 1
a 2m^1
a 2n 1
Q16. Let a , a , ..., a 1 2 30 be an AP,
30
i i 1
S a
= (^) ∑ and ( )
15
2i 1. i 1
T a (^) −
= (^) ∑ If a 5 = 27 and
S − 2T = 75 , then a 10 is equal to
(A) 57 (B) 47 (C) 42 (D) 52
Ans (D)
11. Sequence & Series
Sol: 1 ( )
S 2a 30 1 d 2
S = (^) [ 2a 1 + 29d] 15 …(1)
T = a 1 + a 3 + a 5 + ... +a 29
1 (^ )
2a 15 1 2d 2
2T = 15 2a [ 1 + 28d] …(2)
S - 2T = 15(d) = 75 ⇒ d = 5
Also a 5 = 27
⇒ a 1 + 4d = 27 ⇒ a 1 + 4 5( ) = 27
a 1 = 7
a 10 = a 1 + 9d = 7 + 9 5 ( ) = 52
Q17. If a,b,c are in AP then prove that b + c, c + a, a + b are also in AP.
Sol. ∵ a, b, c in AP
⇒ -a, -b, -c in AP
add (a + b + c) in all terms
⇒ (a + b + c) -a, (a + b + c)-b, (a + b + c)- c in AP
⇒ b + c, c + a, a + b in AP
Q18. If a, b, c are in AP then prove that ( ) ( ) ( )
(^2 2 2 2 ) b + c – a , c +a −b , a +b −c
are also in AP
Sol: ∵ a, b, c in AP
⇒ -2a, -2b , -2c in AP
Add (a + b + c)
(b + c)-a, (c + a)-b, (a + b)-c in AP
Multiply by a + b + c
(^2 2 2 2 ) b + c − a , c + a − b , a + b − c in AP
Q19. If
2 2 2 a ,b , c are in AP then prove that
b + c c + a a + b
are in AP.
Sol: ∵
2 2 2 a ,b , c in AP
add (ab + bc + ca) to each term
⇒ (a + b)(c + a), (a + b)(b + c), (b + c)(c + a) in AP
Divide by (a+b)(b+c)(c+a)
b + c c + a a + b
in AP
13. Sequence & Series
Q1. If 101 arithmetic means & are inserted between 1 and 99 then find
their sum.
Sol: sum of all arithmetic means
= 101 × (mean of 1 and 99)
×
= 101 × 50 = 5050
Q2. Insert 20 AM’s between 4 and 67.
Sol: 4, A 1 , A 2 , A 3 , …, A 20 , 67 in AP
common difference
, d 3 20 1 21
A
1 = 4 + d = 7
A 2 = 4+2d =
A 20 = 4 + 20d = 64
Q3. If p arithmetic means are inserted between 5 and 41 so that the ratio
3
p 1
A 2
A − 5
= then find the value of p.
Sol: 5, A 1 , A 2 , A 3 , …, Ap, 41
common difference
d p 1 p 1
3
p 1
A 2
A − 5
= ⇒ 5A 3 =2Ap (^) − 1
⇒ 5 5( + 3d) = 2 5( + (^) ( p −1 d) )
⇒ 25 + 15d = 10 + ( 2p −2 d)
⇒ ( 2p − 17 d) = 15
2p 17 15 p 1
⇒ 12 2p( − 17 ) = 5 p( + 1 )
⇒ 19p = 209 ⇒ p = 11
Q4. A number sequence a , a , a , ..., a 1 2 3 n is such that
a 1 = 0 ; a 2 = a 1 + 1 ; a 3 = a 2 + 1 ; ... an = an (^) − 1 + 1
Prove that the arithmetic mean of a , a , ..., a 1 2 n is not less than
Sequence & Series 14.
Sol: a 2 = a 1 + 1
on squaring
2 2 a 2 = a 1 + 2a 1 + 1
similarly,
2 2 a 3 = a 2 + 2a 2 + 1
2 2 a 4 = a 3 + 2a 3 + 1
2 2 a n (^) + 1 = an + 2an + 1
on adding all equations, we get
2 2 a n (^) + 1 = a 1 + 2 a 1 + a 2 + ... + an +n
⇒ 2 a( 1 + a 2 + … + an )+ n ≥ 0
a 1 a 2 ... an 1
n 2
≥ H.P.
General Illustration on A.P.
Q5. If (^) ( ) x x 3 3 3
log 2,log 2 5 & log 2 2
are in AP determine x.
Sol: (^) ( )
x x 3 3 3
2log 2 5 log 2 log 2 2
⇒ (^) ( )
x 2 x 3 3
log 2 5 log 2 2 2
⇒ (^) ( )
x 2 x 7 2 5 2 2 2
let 2 x = t
2 t − 5 = 2t − 7
2 t − 12t + 32 = 0
⇒ ( t − 4 ) ( t − 8 ) = 0
x x 2 = 4 , 2 = 8
⇒ x = 2 (rejected as 2 x
` ⇒ x =
Sequence & Series 16.
product of roots = α(α - β)(α + β) =
2(4 - β^2 ) =
⇒ 4 - β 2 =
⇒ β 2 =
β= ±
roots are
for AP
a , d 2 2
n^ (^ )
1 5 5n 6 T n 1 2 2 2
Q10. If the sum of the roots of the equation
2 ax + bx + c = 0 is equal to the
sum of the square of their reciprocals, then show that
2 2 2 bc , ca , ab are
in AP.
Sol: Let roots α β,
now given that 2 2
α + β = + α β
( )
2 2
2
α + β ⇒ α + β = αβ
( ) ( ) ( )
2 2 ⇒ α + β αβ = α + β − 2 αβ
2 2 b c b 2c . a a a a
2 2 2 2 2 3 2
bc b 2c bc ab 2a c a a a
2 2 2 ⇒ ab + bc =2ca
hence
2 2 2 bc , ca , ab are in AP
Q11. Given a , a , a , ..., a 1 2 3 n are in AP. Prove that
1 n 2 n 1 3 n 2 n 1 1 n 1 2 n
a a a a (^) − a a (^) − a a a a a a a
17. Sequence & Series
Sol: We know that a 1 + an = a 2 + an (^) − 1 = a 3 + an 2 (^) − = ... = k(let)
now LHS is
1 n 2 n 1 3 n 2 n 1
1 k k k k ... k a a a a (^) − a a (^) − a a
+^ +^ +^ +
n 1 n 1 2 n 2 3 1 n
k a a a (^) − a a (^) − a a a
+^ +^ +^ +^ +^ +^ +^ +
1 2 n 1 n 1 2 n
2 ... RHS
k a a a a a a a a
+^ + … +^ =^ +^ +^ +^ =
Q12. Let AP (a; d) denote the set of all the terms of an infinite arithmetic
progression with first term a and common difference d > 0.
If AP 1 ; 3( ) ∩ AP 2; 5( ) ∩ AP 3; 7( ) = AP a; d( ) then a+d equals.
Sol: Common terms of two or more than two APs also form an AP whose
common difference is LCM of common differences of these APs
now
AP(1 ; 3) = {1,4,7,10,13,16,19,22,25,…}
AP(2 ; 5) = {2,7,12,17,22,27,32,37,42,…}
AP(3 ; 7) = {3,10,17,24,31,38,…}
first common terms in all these APs will be 52 hence a = 52
d = LCM {3, 5, 7} = 105
hence
a + d = 52 + 105 = 157
Q13. Let a , a , ..., a 1 2 n be a given AP whose common difference is an integer
and Sn = a 1 + a 2 + ... + an. If a 1 = 1, an = 300 and 15 ≤ n ≤ 50 then the
ordered pair ( Sn − 4 , an − 4 ) is equal to
(A) (2480, 248) (B) (2480,249) (C) (2490,249) (D) (2490,248)
Sol: (D)
an = a 1 + ( n −1 d)
300 = 1+(n-1) d
Hence, (^) [ ]
n 1= 14, 49 d
[ ]
n 1= 14, 49 d
×
clearly d = 13 hence n = 24
19. Sequence & Series
(2) ÷ (1)
r r 8 2
2
a 8 r^1
4
9 9 10
t ar 8 2 64
Q2. If th th p , q and th r terms of a G.P. are x, y and z respectively then prove
that
q r r p p q x .y .z 1
− − −
Sol: Given P 1 Tp x AR x
− = ⇒ = …(1)
q 1 Tq y AR y
− = ⇒ = …(2)
r 1 Tr z AR z
− = ⇒ = …(3)
now (^) ( ) q r p 1 q r^ q r ( p^1 )(^ q r) x AR A R − − −^ − −^ − = =
q r pq pr q r A R − − − + = …(4)
( )
r p q 1 r^ p^ r p ( q^1 )(^ r^ p) y AR A R − − −^ − −^ − = =
r p qr pq r p A R − − − + = …(5)
( )
p q r 1 p^ q^ p q ( r^1 )(^ p^ q) z AR A R − − −^ − −^ − = =
p q pr qr p q A R − − − + = …(6)
(4) × (5) × (6)
q r r p p q x .y .z
− − − =A°R°=
Sum of n terms of a G. P.
2 1 ... − = + + + + n S a ar ar ar
( )
n a 1 r S 1 r
where r ≠ 1,If r = 1 then S = na
If r < 1 and n → ∞ then
a S 1 r
∞
Sequence & Series 20.
IMPORTANT POINTS TO REMEMBER
(i) If each term of a GP be multiplied or divided by the same non-zero
quantity, the resulting sequence is also a GP.
(ii) Any 3 consecutive terms of a GP can be taken as
a , a, ar r
. Any 4
consecutive terms of a GP can be taken as
3 3
a a , , ar, ar r r
and so on.
(iii) If a, b, c are in GP ⇒ b 2 = ac
Q1. The sum of first 3 consecutive terms of a GP is 19 and their product is
- find S n . Also compute S∞ if it exists.
Sol. Let terms are α^ /^ β α αβ,^ ,
Given product =.^.^216
α α αβ = β
⇒ α^3 = 6^3 ⇒ α = 6
also sum ⇒ 19
α
put α = 6 then
⇒ 6 + 6β + 6β^2 = 19β
⇒ 6 β^2 - 13β + 6 = 0
⇒ 6 β 2
⇒ 3 β(2β-3) - 2(2β-3) = 0
⇒ (3β-2)(2β-3)=
β = or
hence numbers
for
β = ⇒ (^) ⇒
for
β = ⇒ (^) ⇒
Case–I: GP 9, 6, 4, …
A = 9, r = 2/ n
n
n
S 9 27 1
^
^
