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Home » Find the sum of the integers between 100 and 200 that are (i) divisible by 9 (ii) not divisible by 9 [Hint (ii): These numbers will be: Total numbers – Total numbers divisible by 9]
Find the sum of the integers between 100 and 200 that are
(i) divisible by 9
(ii) not divisible by 9
[Hint (ii): These numbers will be: Total numbers – Total numbers divisible by 9]
Arithmetic Progression | CBSE | Class 10 | Exercise 5.4 | Maths | NCERT Exemplar
Find the sum of the integers between 100 and 200 that are
(i) divisible by 9
(ii) not divisible by 9
[Hint (ii): These numbers will be: Total numbers – Total numbers divisible by 9]

Solution:

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

Let n be the number of terms that are divisible by 9 and are between 100 and 200.

{{a}_{n}}~=\text{ }a\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)d

198\text{ }=\text{ }108\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)9

90\text{ }=\text{ }\left( n\text{ }\text{ }1 \right)9

n1\text{ }=\text{ }10

n\text{ }=\text{ }11

The sum of an AP = {{S}_{n~}}=\text{ }\left( n/2 \right)\text{ }[\text{ }a\text{ }+\text{ }{{a}_{n}}]

{{S}_{n}}~=\text{ }\left( 11/2 \right)\text{ }\times \text{ }\left[ 108\text{ }+\text{ }198 \right]

=\text{ }\left( 11/2 \right)\text{ }\times \text{ }306

=\text{ }11\left( 153 \right)

=\text{ }1683

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

Sum, S\text{ }=\text{ }{{S}_{1}}~\text{ }{{S}_{2}}

Provided here,

{{S}_{1}}~=\text{ }sum\text{ }of\text{ }AP\text{ }101,\text{ }102,\text{ }103,\text{ }\text{ }\text{ }\text{ },\text{ }199

{{S}_{2}}~=\text{ }sum\text{ }of\text{ }AP\text{ }108,\text{ }117,\text{ }126,\text{ }\text{ }\text{ }\text{ },\text{ }198

For the given  AP 101, 102, 103, – – – , 199

The first term, a\text{ }=\text{ }101

The common difference, d\text{ }=\text{ }199

No. of terms = n

Therefore,

{{a}_{n}}~=\text{ }a\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)d

199\text{ }=\text{ }101\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)1

98\text{ }=\text{ }\left( n\text{ }\text{ }1 \right)

n\text{ }=\text{ }99

The sum of an AP = {{S}_{n~}}=\text{ }\left( n/2 \right)\text{ }[\text{ }a\text{ }+\text{ }{{a}_{n}}]

The sum of this AP,

{{S}_{1~}}=\text{ }\left( 99/2 \right)\text{ }\times \text{ }\left[ 199\text{ }+\text{ }101 \right]

=\text{ }\left( 99/2 \right)\text{ }\times \text{ }300

=\text{ }99\left( 150 \right)

=\text{ }14850

For the given AP 108, 117, 126, – – – – , 198

The first term, a\text{ }=\text{ }108

The common difference, d\text{ }=\text{ }9

The last term, {{a}_{n}}~=\text{ }198

No. of terms = n

Therefore,

{{a}_{n}}~=\text{ }a\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)d

198\text{ }=\text{ }108\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)9

10\text{ }=\text{ }\left( n\text{ }\text{ }1 \right)

n\text{ }=\text{ }11

The sum of an AP = {{S}_{n~}}=\text{ }\left( n/2 \right)\text{ }[\text{ }a\text{ }+\text{ }{{a}_{n}}]

The sum of this AP,

{{S}_{2~}}=\text{ }\left( 11/2 \right)\text{ }\times \text{ }\left[ 108\text{ }+\text{ }198 \right]

=\text{ }\left( 11/2 \right)\text{ }\times \text{ }\left( 306 \right)

=\text{ }11\left( 153 \right)

=\text{ }1683

In the equation, S\text{ }=\text{ }{{S}_{1}}~\text{ }{{S}_{2}},, substituting the value of S1 and S2

S\text{ }=\text{ }{{S}_{1}}~+\text{ }{{S}_{2}}

=\text{ }14850\text{ }\text{ }1683

=\text{ }13167.

i

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