(iit Jee) Sanjay Mishra - Algebra 1 Fundamentals Of Mathematics-pearson.pdf

3. 1O

►

Fundamentals of Mathematics-Algebra I

3. The number of terms common to two APs 3, 7, 11 ,.... , 407 and 2, 9, 16, . , 709 is (a) 21 (b) 28 (c) 14 (d) None of these 4. The largest term common to the sequences 1, 11 , 21, 31 , ..... upto 100 terms and 31 , 36, 41 , 46, ..... upto 100 terms is (a) 381 (b) 471 (c) 281 (d) None of these 5. Let xi' x 2• .. be positive integers in AP, such that x 1 + x 2 + xJ= 12 andx 4 + x 6 = 14. Thenx5 is (a) 7 (b) 1 (c) 4 (d) None of these

9. Ifs r denotes the sum of the first r terms of an AP, then s -s J, ,-i is equal to Sz, - Sz,- 1

10. Let s denote the sum of the first n terms of an AP If s 211 = 3s,,, then sJ11 :s11 is equal to 1l

11. The mu, term of an arithmetic progression is x and the nu, term is y. Then the sum of the first (m + n) terms is

(a) m+n[(x+ y)+ x-yJ 2 m-n

y]

(b) m+n[(x-y)+ x+ 2 m-n

n consecutive terms have the value n, the 150th term is (a) 17 (b) 16 (c) 18 (d) None of these

8. The sum of the first n terms (n > 1) of an AP is 153 and the common difference is 2. If the first term is an integer, the number of possible values of n is (a) 3 (b) 4 (c) 5 (d) 6

(b) 6 (d) 10

(a) 4 (c) 8

6. In the sequence 1, 2, 2, 3, 3, 3, 4, 4, 4, 4 .... ; where

7. Six numbers are in AP such that their sum is 3. The first number is four times the third number. The fifth number is equal to (a) -15 (b) -3 (c) 9 (d) -4

(b) 2r + 1 (d) 2r + 3

(a) 2r - 1 (c) 4r + 1

(c) _!_[x+y + x+y] 2 m+n m-n (d) _!_[x+y _x-y] 2 m+n m-n

s

s

12. Ifs denotes the sum of r terms of an AP and ---f = --1, . a b = c, then Sc lS (a) cJ (b) c/ab (c) abc (d) a + b + c 13. The sum of the series: 12 - 2 2 + 32 - 42 + 52 - 100 2 is (a) -10100 (b) -5050 (c) -2525 (d) None of these

-

62 + .

Answer Keys 1. (c) 11. (a)

2. (b) 12. (a)

3. (c) 13. (b)

4. (d)

5. (a)

6. (a)

7. (d)

8. (c)

9. (b)

10. (b)

Similarly,

Properties of arithmetic progression

n

(t 2 +t"_1)

n

n

2

2

= [a +d +a +(n-2)d] = [2a +(n-l)d]

(a) nu, term can also be written as, t,, = s 11 -s 11_ 1

2

(b) Summation of equidistant terms from beginning and end of an AP is always constant and is equal to sum of first and last term

and so on. Hence the result. (c) If nu, term t11 = an + b, then the series so formed is an AP (with common difference a) Proof: ⇒ t,,+1

t = an + b 1l

= a(n + 1) + b

⇒ t,,+1 - t,, ⇒

= a constant independent of n.

given sequence is an AP

Sequence and Progression

(d) If sum of first n terms of a series sn = an 2 + bn + c, then series so formed is an AP (with common difference 2a) provided (c = 0). Proof: t 1 = S1 =a+ b + c t2 = s2 -SI= (4a + 2b + c)-(a + b+ c) = 3a + b t3 = S 3 -S2 = (9a + 3b + c)-(4a + 2b + c) =Sa+ b Now, tn = S,, - S,,_ 1 = a[n 2 - (n - 1)2] + b[n - (n - l)] = a(2n - 1) + b; n ~ 2 ⇒ t t,,_ 1 = a[2n - 1] - a[2(n - 1) - 1] = 2a for n ~ 3 ⇒ t , t , t , ..... are in AP. But t - t = 2a -c 2 3 4 2 1 ti' t 2 , t 3 , t 4 , .... , will be in AP Provided c = 0 -

11

(e) If every term of an AP is increased or decreased by the same quantity, the resulting terms will also be in AP with no change in common difference. Proof: Let the AP be a, a + d, a + 2d, a + 3d, ... Let each term be changed by ± k, then new sequence is a ± k, a + d ± k, a + 2d ± k, ...... which is again an AP with common difference d. (f) If every term of an AP (C.D. = d) is multiplied or divided by the same non-zero quantity k, then resulting terms will be in AP with new common difference equal to d.k or d/k. Proof: On multiplying each term by k or 1/k we get ka, k(a + d), k(a + 2d), k(a + 3d), ... or a/k, (a+d)/k., (a+ 2d)/k, .... which are clearly in AP with common difference kd and d/k respectively.

(g) is an AP ⇒ is also an AP VkE 1, 2, ... , n with common difference= (-d) (h) and are two APs, then is also an AP with common difference = da ±db. (i) If equal number of terms (say k terms of an AP) are dropped alternately, the resulting terms lie in AP with common difference = (k + 1) d. e.g. a, a + d, a + 2d, a + 3d, a + 4d, a + 5d, a + 6d, then for sequence a, a, 3d, a + 6d, number of terms dropped = 2; so common difference = (2 + 1) d= 3d (j) If equal number of terms say 'k' terms of an AP are grouped together and sum of terms in each group is obtained; then the sums are in AP with common difference k2d e.g., a + a + d + a + 2d, a + 3d + a + 4d + a + 5d,

Proof: s

+1

meanings)

2n+l = - [a 1 + a 2

2

(terms have their usual

n+ ] 1

2

= (2n + 1) a

11

+ 1=

(/) If terms a 1, a 2,

(2n + 1) (middle term)

••. ,

a

2

a

n- , 1

211

are in AP The sum of

these terms will be equal to (2n) ( a,, + an+i ), where 2 a,, + a,,+i = AM of middle terms. 2

2n 2n s, = [a 1 +a,]= - [2a 1 + (2n - l)d] (terms 2 2 have their usual meanings) 2n 2n = - [{a +(n-l)d}+{a +nd}]=-[a +a ]

Proof:

Oil

.!.fl

2

~ (:n) [ a'. +2a•., ] ~ 2n x (~M of two mi~lot::,)

(m) If for two different AP's ratio of sum of 1st n terms is S,, = f(n) then ~ = f(2n -1) ¢(n)' t',, ¢(2n-1)

s; Proo: f

⇒

.) (1

s,,

a·

1ven ~ =

f(n)

¢(n)

%[ 2a+(n-l)d]

f(n)

_'l_[ 2a'+ (n -1) d']

¢(n)

2

⇒

11

a+(n-l)d/2

f(n)

a'+(n-l)d'/2

¢(n)

n-1 Put - - = (m -1) ⇒ n = 2m -1 2

a+(m-l)d

f(2m-1)

-----= ----

a '+ (m -

I) d '

¢ ( 2m -

l)

Tm

f(2m-l)

Tm '

¢(2m-1)

⇒

T f (2n-1) In general -" = - - ' T,, ' ¢ ( 2n -1)

(n) If for two AP's, ratio of nth term is given by T,,

F;(n)

S,,

F;(n;l)

(n;1)

-=-- ⇒ -=----

T:

F 2 (n)

s:

F 2

a + 6d + a + 7d + a + 8d S 3 ::::3a+21d

⇒

S 1, S 2 , S 3 , ... are in AP with common difference of sums = 9d i.e., (3 )2d = 9d. (k) If terms a 1, a 2 , .•. , a,,, a + 1, .••• , a + 1 are in AP Then sum of these terms will be equal to (2n + 1)a + 1. Here total number of terms in the series is (2n + 1) and middle term is a,,+1· 11

211

11

3.11

2 = ( n+l) [2a 1 +(2n)d]=(2n+ l)[a 1 +nd]

11

S 1 =3a+3J

211

«:

Proof:

Given!.-:.._= a+(n-l)d T,,' a'+(n-l)d' n

-[2a + (n - l)d] ⇒ _n = ~ 2 ~ - - - S' n ,, -[2a'+(n-l)d'] 2 S

F;(n) F2 (n)

a+(T)d

3. 12

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Fundamentals of Mathematics-Algebra I

POINTS TO REMEMBER (i) 3 numbers in AP be taken as

a - d, a, a+ d

[c.d. =dl

(ii) 4 numbers in AP be taken as

a - 3d, a - d, a+ d, a + 3d

(iii) 5 numbers in AP be taken as

a - 2d, a - d, a, a+ d, a + 2d

[c.d=2dJ [c.d. =dl

NOTE In general, if we have to take (2r + 1) terms in AP, we take them as a - rd, a - (r - 7)d, ..., a - d, a, a+ d, .., a+ rd and if we have to take 2r terms in AP, we take them as (a - (2r- 1 )d), (a - (2r- 3)d), ..... ,(a+ (2r- 3)d), (a+ (2r- 1 )d).

ILLUSTRATION 16: If the roots of the equation x3

-- l 2x2 + 39x -- 28 = 0 are in AP, then find its common difference.

+d

SOLUTION: Let roots be a, {3, y and a = a -- d, {3 = a, y = a ⇒ a=

then a+ {3 +y = 3a = -{--12)

a{3y = a(a2 -d'-) = -{--28)

⇒

d

=

4

±3

ILLUSTRATION 17: If the ratio of the sum of n terms of two AP's is (3n

+ 1): (2n + 3), then find the ratio of their

11th term. SOLUTION: Here (Sn)1

= 3n+l ⇒ n/2[2lli_ +(n-l)d1 ]

(Sn) 2 ⇒

⇒

2n+3

n/2[2a2 +(n-I)d2]

2lJi +(n- l)d1

3n+l

2a2 + (n- l)d2

2n+3

3n+l 2n+3

(n--1) d lli + - ~ 2 I 3n+l (n-I) d 2n+3 a2+-~ 2 2 i;l = lli + lOdl 7;1' a2 +IOd2 n -- 1 = 10 2 lJi +10d1

(i)

(ii)

⇒ n = 21; putting the values of n in (i),

a 2 +IOd2

3x21+1 2x21+3

64

45

ILLUSTRATION 18: The sum of four integers in AP is 24 and their product is 945. Find the numbers. SOLUTION: Let the numbers are a -- 3d, a -- d, a

+ d, a + 3d

given a -- 3d + a -- d + a + d + a + 3d = 24 or 4a = 24

. ·. a = 6 and(a -- 3d) (a -d) (a + d; (a + 3d) = 945

⇒

(a2

⇒

4

--

9d 2) (a 2

2

--

d -40d + 144

d =

2

=

945

105

⇒

)

⇒

(36 -- 9d 2 ) (36 -
d -40d 2 + 39 4

=

0

=

945

Sequence and Progression

«:

3.13

or (d 2 -l)(d 2 -39)=0 Since numbers are integers d

2

;t:39 ⇒ d 2 -l =O ⇒ d=±l

Hence the four integers are 3, 5, 7, 9 or 9, 7, 5, 3 ILLUSTRATION19: If b+c-a c+a-b a+b-c areinAP,thenshowthatl/a, lib, llcarealsoinAP.

a

'

b

c

'

b+c-a c+a-b a+b-c

• AP .

SOLUTION: - - - - - - - - - , are 1n

a

'

b

c

'

b+c-a c+a-b a+b-c . . . - - - + 2, - - - + 2 , - - - + 2 are m AP (addmg 2 m each term) a b c a+b+c c+a+b a+b+c . AP or - - - , - - - , - - - , are m . a b c (dividing by (a + b + c) each term) lla, lib, lie are in AP ILLUSTRATION 20: (a) If a 2

+ ab + be + ca, b2 + ab + be + ca, c2 + ab + be + ca are in AP, then show that

1 1 1 . - - - - - - are m AP b+c' c+a' a+b . AP, then prove th at - 1- , -1- , -1- are m . AP (b) If -a- , -b- , -c- are m b+c c+a a+b b+c c+a a+b ⇒

+ ab + be + ca, b2 + ab + be + ca, c 2 + ab + be + ca are in AP (a + b)(a + c), (b + c)(b + a), (c + a)(c + b) are in AP

⇒

- - , - - , - - are in AP (dividing each term by (a + b) (b + c) (c + a))

SOLUTION: (a) a 2

1

1

1

b+c c+a a+b

Hence proved.

(b) _a___ b ___ c_ are in AP. b+c 'c+a' a+b

⇒ ⇒

⇒

_a_+l,-b-+1,-c-+1 are in AP (adding 1 to each term) b+c c+a a+b

a+b+c a+b+c a+b+c are in AP. b+c ' c+a ' a+b 1 1 1 . . . . - - , - - , - - are m AP] (div1dmg each term by (a + b + c)) b+c c+a a+b Hence proved.

ILLUSTRATION 21: The sequence al' a 2 , a 3 , . . .... . a98 satisfies the relation an+1 =an + l for n = l, 2, 3, ..... .97 and has 49

the sum equal to 4949. then, evaluate

I,a

21 .

1~1

SOLUTION: Given recurrence equation is an+ 1 = an ⇒

a2 = a 1 + 1; a3 = a 2 + 1 and so on. the series

⇒

⇒

+l

is ⇒

a 1 + (a 1 + 1) + (a 1 + 1 + l)+.. ..... ..(a1 + 97)

98

4949 = - 2 (a I + a I + 97) 2a1 = 101 - 97

⇒

a 1 = 2, a 2 = 3, a 4 = 5, and so on

Required sum= a 2 + a 4 + .... . + a 98 = 3 + 5 + 7 ... ..+ 99 49 S = -(3+99) = 49 n 2

X

51 = 2499