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1.

Find the sum of all integers between 84 and 719, which are multiples of 5

Answer»

The series is 85, 90, 95, …, 715

Let there be ‘n’ terms in the AP

So, a = 85, d = 90-85 = 5, an = 715

an = a + (n - 1)d

715 = 85 + (n - 1)5

715 = 85 + 5n – 5

5n = 715 – 85 + 5

5n = 635

n = 635/5

= 127

By using the formula,

Sum of n terms, S = n/2 [a + l]

= 127/2 [85 + 715]

= 127/2 [800]

= 127 [400]

= 50800

∴ The sum of all integers between 84 and 719, which are multiples of 5 is 50800.
Discussion
2.

Three numbers are in A.P. If the sum of these numbers be 27 and the product 648, find the numbers

Answer»

Given: 

Sum of first three terms is 27

Let us assume the first three terms as a – d, a, a + d [where a is the first term and d is the common difference]

So, sum of first three terms is

a – d + a + a + d = 27

3a = 27

a = 9

It is given that the product of three terms is 648

So, a3 – ad2 = 648

Substituting the value of a = 9, we get

93 – 9d2 = 648

729 – 9d2 = 648

81 = 9d2

d = 3 or d = – 3

Hence, the given terms are a – d, a, a + d which is 6, 9, 12.
Discussion
3.

If x, y, z are in A.P. and A1is the A.M. of x and y, and A2 is the A.M. of y and z, then prove that the A.M. of A1 and A2 is y.

Answer»

Given that,

A1 = AM of x and y

And A2 = AM of y and z

So, A1 = (x + y)/2

A2 = (y + x)/2

AM of A1 and A2 = (A1 + A2)/2

= [(x + y)/2 + (y + z)/2]/2

= [x + y + y + z]/2

= [x + 2y + z]/2

Since x, y, z are in AP, y = (x + z)/2

AM = [(x + z/2) + (2y/2)]/2

= (y + y)/2

= 2y/2

= y

Hence proved.
Discussion
4.

A farmer buys a used tractor for ₹ 12000. He pays ₹ 6000 cash and agrees to pay the balance in annual instalments of ₹ 500 plus 12% interest on the unpaid amount. How much the tractor cost him?

Answer»

Given: Price of the tractor is ₹12000.

We need to find the total cost of the tractor if he buys it in installments.

Total price = ₹ 12000

Paid amount = ₹ 6000

Unpaid amount = ₹ 12000 – 6000 = ₹ 6000

He pays remaining ₹ 6000 in ‘n’ number of installments of ₹ 500 each.

So, n = 6000/500 = 12

Cost incurred by him to pay remaining 6000 is

The AP will be:

(500 + 12% of 6000) + (500 + 12% of 5500) + … up to 12 terms

500 × 12 + 12% of (6000 + 5500 + … up to 12 terms)

By using the formula,

Sn = n/2 [2a + (n – 1)d]

n = 12, a = 6000, d = -500

S12 = 500×12 + 12/100 × 12/2 [2(6000) + (12 - 1) (-500)]

= 6000 + 72/100 [12000 + 11 (-500)]

= 6000 + 72/100 [12000 – 5500]

= 6000 + 72/100 [6500]

= 6000 + 4680

= 10680

Total cost = 6000 + 10680

= 16680

∴ The total cost of the tractor if he buys it in installment is ₹ 16680.
Discussion
5.

Find the sum of the following series:(i) 2 + 5 + 8 + … + 182(ii) 101 + 99 + 97 + … + 47(iii) (a – b)2 + (a2 + b2) + (a + b)2 + s…. + [(a + b)2 + 6ab]

Answer»

(i) 2 + 5 + 8 + … + 182

First term, a = a1 = 2

Common difference, d = a2 – a1 = 5 – 2 = 3

an term of given AP is 182

an = a + (n - 1)d

182 = 2 + (n - 1)3

182 = 2 + 3n – 3

182 = 3n – 1

3n = 182 + 1

n = 183/3

= 61

Now,

By using the formula,

S = n/2 (a + l)

= 61/2 (2 + 182)

= 61/2 (184)

= 61 (92)

= 5612

∴ The sum of the series is 5612

(ii) 101 + 99 + 97 + … + 47

First term, a = a1 = 101

Common difference, d = a2 – a1 = 99 – 101 = -2

an term of given AP is 47

an = a + (n - 1)d

47 = 101 + (n - 1)(-2)

47 = 101 – 2n + 2

2n = 103 – 47

2n = 56

n = 56/2 = 28

Then,

S = n/2 (a + l)

= 28/2 (101 + 47)

= 28/2 (148)

= 14 (148)

= 2072

∴ The sum of the series is 2072

(iii) (a – b)2 + (a2 + b2) + (a + b)2 + s…. + [(a + b)2 + 6ab]

First term, a = a1 = (a - b)2

Common difference, d = a2 – a1 = (a2 + b2) – (a – b)2 = 2ab

an term of given AP is [(a + b)2 + 6ab]

an = a + (n - 1) d

[(a + b)2 + 6ab] = (a - b)2 + (n - 1)2ab

a2 + b2 + 2ab + 6ab = a2 + b2 – 2ab + 2abn – 2ab

a2 + b2 + 8ab – a2 – b2 + 2ab + 2ab = 2abn

12ab = 2abn

n = 12ab / 2ab

= 6

Then,

S = n/2 (a + l)

= 6/2 ((a - b)2 + [(a + b)2 + 6ab])

= 3 (a2 + b2 – 2ab + a2 + b2 + 2ab + 6ab)

= 3 (2a2 + 2b2 + 6ab)

= 3 × 2 (a2 + b2 + 3ab)

= 6 (a2 + b2 + 3ab)

∴ The sum of the series is 6 (a2 + b2 + 3ab)
Discussion
6.

If a, b, c are in A.P., then show that:(i) a2(b + c), b2(c + a), c2(a + b) are also in A.P.(ii) b + c – a, c + a – b, a + b – c are in A.P.(iii) bc – a2, ca – b2, ab – c2 are in A.P.

Answer»

(i) a2(b + c), b2(c + a), c2(a + b) are also in A.P.

If b2(c + a) – a2(b + c) = c2(a + b) – b2(c + a)

b2c + b2a – a2b – a2c = c2a + c2b – b2a – b2c

Given, b – a = c – b

And since a, b, c are in AP,

c(b2 – a2 ) + ab(b – a) = a(c2 – b2 ) + bc(c – b)

(b – a) (ab + bc + ca) = (c – b) (ab + bc + ca)

Upon cancelling, ab + bc + ca from both sides

b – a = c – b

2b = c + a [which is true]

Hence, given terms are in AP

(ii) b + c – a, c + a – b, a + b – c are in A.P.

If (c + a – b) – (b + c – a) = (a + b – c) – (c + a – b)

Then, b + c – a, c + a – b, a + b – c are in A.P.

Let us consider LHS and RHS

(c + a – b) – (b + c – a) = (a + b – c) – (c + a – b)

2a – 2b = 2b – 2c

b – a = c – b

And since a, b, c are in AP,

b – a = c – b

Hence, given terms are in AP.

(iii) bc – a2, ca – b2, ab – c2 are in A.P.

If (ca – b2) – (bc – a2) = (ab – c2) – (ca – b2)

Then, bc – a2, ca – b2, ab – c2 are in A.P.

Let us consider LHS and RHS

(ca – b2) – (bc – a2) = (ab – c2) – (ca – b2)

(a – b2 – bc + a2) = (ab – c2 – ca + b2)

(a – b) (a + b + c) = (b – c) (a + b + c)

a – b = b – c

And since a, b, c are in AP,

b – c = a – b

Hence, given terms are in AP
Discussion
7.

If (b + c)/a, (c + a)/b, (a + b)/c are in AP., prove that:(i) 1/a, 1/b, 1/c are in AP(ii) bc, ca, ab are in AP

Answer»

(i) 1/a, 1/b, 1/c are in AP

If 1/a, 1/b, 1/c are in AP then,

1/b – 1/a = 1/c – 1/b

Let us consider LHS:

1/b – 1/a = (a - b)/ab

= c(a - b)/abc [by multiplying with ‘c’ on both the numerator and denominator]

Let us consider RHS:

1/c – 1/b = (b - c)/bc

= a(b - c)/bc [by multiplying with ‘a’ on both the numerator and denominator]

Since, (b + c)/a, (c + a)/b, (a + b)/c are in AP

a(b - c) = c(a - b)

LHS = RHS

Hence, the given terms are in A.P

(ii) bc, ca, ab are in AP

If bc, ca, ab are in AP then,

ca – bc = ab – ca

c (a - b) = a (b - c)

If 1/a, 1/b, 1/c are in AP then,

1/b – 1/a = 1/c – 1/b

c (a - b) = a (b - c)

Hence, the given terms are in AP
Discussion
8.

If a(1/b + 1/c), b(1/c + 1/a), c(1/a + 1/b) are in AP., prove that a, b, c are in AP.

Answer»

Here, we know a(1/b + 1/c), b(1/c + 1/a), c(1/a + 1/b) are in AP

Also, a(1/b + 1/c) + 1, b(1/c + 1/a) + 1, c(1/a + 1/b) + 1 are in AP

Let us take LCM for each expression then we get,

(ac + ab + bc)/bc , (ab + bc + ac)/ac, (cb + ac + ab)/ab are in AP

1/bc, 1/ac, 1/ab are in AP

Let us multiply numerator with ‘abc’, we get

abc/bc, abc/ac, abc/ab are in AP

∴ a, b, c are in AP.

Hence proved.
Discussion
9.

If a, b, c are in A.P., prove that:(i) (a – c)2 = 4 (a – b) (b – c)(ii) a2 + c2 + 4ac = 2 (ab + bc + ca)(iii) a3 + c3 + 6abc = 8b3

Answer»

(i) (a – c)2 = 4 (a – b) (b – c)

Let us expand the above expression

a2 + c2 – 2ac = 4(ab – ac – b2 + bc)

a2 + 4c2b2 + 2ac – 4ab – 4bc = 0

(a + c – 2b)2 = 0

a + c – 2b = 0

Since a, b, c are in AP

b – a = c – b

a + c – 2b = 0

a + c = 2b

Hence, (a – c)2 = 4 (a – b) (b – c)

(ii) a2 + c2 + 4ac = 2 (ab + bc + ca)

Let us expand the above expression

a2 + c2 + 4ac = 2 (ab + bc + ca)

a2 + c2 + 2ac – 2ab – 2bc = 0

(a + c – b)2 – b2 = 0

a + c – b = b

a + c – 2b = 0

2b = a + c

b = (a + c)/2

Since a, b, c are in AP

b – a = c – b

b = (a + c)/2

Hence, a2 + c2 + 4ac = 2 (ab + bc + ca)

(iii) a3 + c3 + 6abc = 8b3

Let us expand the above expression

a3 + c3 + 6abc = 8b3

a3 + c3 – (2b)3 + 6abc = 0

a3 + (-2b)3 + c3 + 3a(-2b)c = 0

Since, if a + b + c = 0, a3 + b3 + c3 = 3abc

(a – 2b + c)3 = 0

a – 2b + c = 0

a + c = 2b

b = (a + c)/2

Since a, b, c are in AP

a – b = c – b

b = (a + c)/2

Hence, a3 + c3 + 6abc = 8b3
Discussion