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Find the sum of the following APS.
(i) 0.6,1.7,2.8, \ldots \ldots \ldots, to 100 terms
(ii) 1 / 15,1 / 12,1 / 10, \ldots \ldots, to 11 terms
Arithmetic Progression | CBSE | Class 10 | Maths | NCERT
Find the sum of the following APS.
(i) 0.6,1.7,2.8, \ldots \ldots \ldots, to 100 terms
(ii) 1 / 15,1 / 12,1 / 10, \ldots \ldots, to 11 terms

Solutions:

(i) Provided that, 0.6,1.7,2.8, \ldots, to 100 terms

And for this A.P.,

The first term, a=0.6

And the common difference, d=a_{2}-a_{1}=1.7-0.6=1.1

We all know that,for the sum of nth term in the AP series the formula is,

S_{n}=n / 2[2 a+(n-1) d] \mathrm{S}_{12}=50 / 2[1.2+(99) \times 1.1]

=50[1.2+108.9]

=50[110.1]

=5505

(ii) Provided that, 1 / 15,1 / 12,1 / 10, \ldots \ldots, to 11 terms

And for this A.P.,

The first term, a=1 / 5

And the common difference, d=a_{2}-a_{1}=(1 / 12)-(1 / 5)=1 / 60

And the number of terms \mathrm{n}=11

We all know that,for the sum of nth term in the AP series the formula is,

S_{n}=n / 2[2 a+(n-1) d]

S n=\frac{11}{2}\left[2\left(\frac{1}{15}\right)+\frac{(11-1) 1}{60}\right]

=11 / 2(2 / 15+10 / 60)

=11 / 2(9 / 30)

=33 / 20

i

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