Chapter 5: Arithmetic Progressions

We know that, in an A.P.,

First term = a

Common difference = d

Number of terms of an AP = n

According to the question,

We have,

S₅ + S₇ = 167

Using the formula for sum of n terms,

S_n = (n/2) [2a + (n-1)d]

So, we get,

(5/2) [2a + (5-1)d] + (7/2)[2a + (7-1)d] = 167

5(2a + 4d) + 7(2a + 6d) = 334

10a + 20d + 14a + 42d = 334

24a + 62d = 334

12a + 31d = 167

12a = 167 – 31d …(1)

We have,

S₁₀ = 235

(10/2) [2a + (10-1)d] = 235

5[ 2a + 9d] = 235

2a + 9d = 47

Multiplying L.H.S and R.H.S by 6,

We get,

12a + 54d = 282

From equation (1)

167 – 31d + 54d = 282

23d = 282 – 167

23d = 115

d = 5

Substituting the value of d = 5 in equation (1)

12a = 167 – 31(5)

12a = 167 – 155

12a = 12

a = 1

We know that,

S₂₀ = (n/2) [2a + (20 – 1)d]

= 20/(2[2(1) + 19 (5)])

= 10[ 2 + 95]

= 970

Therefore, the sum of first 20 terms is 970.

(i) Sum of those integers between 1 and 500 which are multiples of 2 as well as of 5.

We know that,

Multiples of 2 as well as of 5 = LCM of (2, 5) = 10

Multiples of 2 as well as of 5 between 1 and 500 = 10, 20, 30…, 490.

Hence,

We can conclude that 10, 20, 30…, 490 is an AP with common difference, d = 10

First term, a = 10

Let the number of terms in this AP = n

Using n^th term formula,

a_n = a + (n – 1)d

490 = 10 + (n – 1)10

480 = (n – 1)10

n – 1 = 48

n = 49

Sum of an AP,

S_n = (n/2) [a + a_n], here a_n is the last term, which is given]

= (49/2) × [10 + 490]

= (49/2) × [500]

= 49 × 250

= 12250

Therefore, sum of those integers between 1 and 500 which are multiples of 2 as well as of 5 = 12250

(ii) Sum of those integers from 1 to 500 which are multiples of 2 as well as of 5.

We know that,

Multiples of 2 as well as of 5 = LCM of (2, 5) = 10

Multiples of 2 as well as of 5 from 1 and 500 = 10, 20, 30…, 500.

Hence,

We can conclude that 10, 20, 30…, 500 is an AP with common difference, d = 10

First term, a = 10

Let the number of terms in this AP = n

Using n^th term formula,

a_n = a + (n – 1)d

500 = 10 + (n – 1)10

490 = (n – 1)10

n – 1 = 49

n = 50

Sum of an AP,

S_n = (n/2) [ a + a_n], here a_n is the last term, which is given]

= (50/2) ×[10+500]

= 25× [10 + 500]

= 25(510)

= 12750

Therefore, sum of those integers from 1 to 500 which are multiples of 2 as well as of 5= 12750

(iii) Sum of those integers from 1 to 500 which are multiples of 2 or 5.

We know that,

Multiples of 2 or 5 = Multiple of 2 + Multiple of 5 – Multiple of LCM (2, 5)

Multiples of 2 or 5 = Multiple of 2 + Multiple of 5 – Multiple of LCM (10)

Multiples of 2 or 5 from 1 to 500 = List of multiple of 2 from 1 to 500 + List of multiple

of 5 from 1 to 500 – List of multiple of 10 from 1 to 500

= (2, 4, 6… 500) + (5, 10, 15… 500) – (10, 20, 30… 500)

Required sum = sum(2, 4, 6,…, 500) + sum(5, 10, 15,…, 500) – sum(10, 20, 30,., 500)

Consider the first series,

2, 4, 6, …., 500

First term, a = 2

Common difference, d = 2

Let n be no of terms

a_n = a + (n – 1)d

500 = 2 + (n – 1)²

498 = (n – 1)²

n – 1 = 249

n = 250

Sum of an AP, S_n = (n/2) [ a + a_n]

Let the sum of this AP be S_1,

S₁ = S₂₅₀ = (250/2) ×[2+500]

S₁ = 125(502)

S₁ = 62750 … (1)

Consider the second series,

5, 10, 15, …., 500

First term, a = 5

Common difference, d = 5

Let n be no of terms

By nth term formula

a_n = a + (n – 1)d

500 = 5 + (n – 1)

495 = (n – 1)5

n – 1 = 99

n = 100

Sum of an AP, S_n = (n/2) [ a + a_n]

Let the sum of this AP be S_2,

S₂ = S₁₀₀ = (100/2) ×[5+500]

S₂ = 50(505)

S₂ = 25250 … (2)

Consider the third series,

10, 20, 30, …., 500

First term, a = 10

Common difference, d = 10

Let n be no of terms

a_n = a + (n – 1)d

500 = 10 + (n – 1)10

490 = (n – 1)10

n – 1 = 49

n = 50

Sum of an AP, S_n = (n/2) [ a + a_n]

Let the sum of this AP be S_3,

S₃ = S₅₀ = (50/2) × [2+510]

S₃ = 25(510)

S₃ = 12750 … (3)

Therefore, the required Sum, S = S₁ + S₂ – S₃

S = 62750 + 25250 – 12750

= 75250

We know that,

First term of an AP = a

Common difference of AP = d

n^th term of an AP, a_n = a + (n – 1)d

According to the question,

a_s = ½ a₂

2a₈ = a₂

2(a + 7d) = a + d

2a + 14d = a + d

a = – 13d …(1)

Also,

a₁₁ = 1/3 a₄ + 1

3(a + 10d) = a + 3d + 3

3a + 30d = a + 3d + 3

2a + 27d = 3

Substituting a = -13d in the equation,

2 (- 13d) + 27d = 3

d = 3

Then,

a = – 13(3)= – 39

Now,

a₁₅  = a + 14d

= – 39 + 14(3)

= – 39 + 42

= 3

So 15^th term is 3.

We know that,

First term of an AP = a

Common difference of AP = d

n^th term of an AP, a_n = a + (n – 1)d

Since, n = 37 (odd),

Middle term will be (n+1)/2 = 19^th term

Thus, the three middle most terms will be,

18^th, 19^th and 20^th terms

According to the question,

a₁₈ + a₁₉ + a₂₀ = 225

Using a_n = a + (n – 1)d

a + 17d + a + 18d + a + 19d = 225

3a + 54d = 225

3a = 225 – 54d

a = 75 – 18d … (1)

Now, we know that last three terms will be 35^th, 36^th and 37^th terms.

According to the question,

a₃₅ + a₃₆ + a₃₇ = 429

a + 34d + a + 35d + a + 36d = 429

3a + 105d = 429

a + 35d = 143

Substituting a = 75 – 18d from equation 1,

75 – 18d + 35d = 143 [ using eqn1]

17d = 68

d = 4

Then,

a = 75 – 18(4)

a = 3

Therefore, the AP is a, a + d, a + 2d….

i.e. 3, 7, 11….

(i) The number between 100 and 200 which is divisible by 9 = 108, 117, 126, …198

Let the number of terms between 100 and 200 which is divisible by 9 = n

a_n = a + (n – 1)d

198 = 108 + (n – 1)9

90 = (n – 1)9

n – 1 = 10

n = 11

Sum of an AP = S_n = (n/2) [ a + a_n]

S_n = (11/2) × [108 + 198]

= (11/2) × 306

= 11(153)

= 1683

(ii) Sum of the integers between 100 and 200 which is not divisible by 9 = (sum of total numbers between 100 and 200) – (sum of total numbers between 100 and 200 which is divisible by 9)

Sum, S = S₁ – S₂

Here,

S₁ = sum of AP 101, 102, 103, – – – , 199

S₂ = sum of AP 108, 117, 126, – – – , 198

For AP 101, 102, 103, – – – , 199

First term, a = 101

Common difference, d = 199

Number of terms = n

Then,

a_n = a + (n – 1)d

199 = 101 + (n – 1)1

98 = (n – 1)

n = 99

Sum of an AP = S_n = (n/2) [ a + a_n]

Sum of this AP,

S₁ = (99/2) × [199 + 101]

= (99/2) × 300

= 99(150)

= 14850

For AP 108, 117, 126, – – – – , 198

First term, a = 108

Common difference, d = 9

Last term, a_n = 198

Number of terms = n

Then,

a_n = a + (n – 1)d

198 = 108 + (n – 1)9

10 = (n – 1)

n = 11

Sum of an AP = S_n = (n/2) [ a + a_n]

Sum of this AP,

S₂ = (11/2) × [108 + 198]

= (11/2) × (306)

= 11(153)

= 1683

Substituting the value of S₁ and S₂ in the equation, S = S₁ – S₂

S = S₁ + S₂

= 14850 – 1683

= 13167

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