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Find the sums given below:
Arithmetic Progression | CBSE | Class 10 | Maths | NCERT
Find the sums given below:

(i) 7+10 \frac{1}{2}+14+\ldots \ldots \ldots \ldots \ldots . .+84

(ii) 34+32+30+\ldots \ldots \ldots . .+10

Solutions:

(i) For the given A.P., 7+10 \frac{1}{2}+14+\ldots \ldots \ldots \ldots \ldots . .+84

The First term, a=7

n^{\text {th }} term, a_{n}=84

The Common difference, \mathrm{d}=\mathrm{a}<em>{2}-\mathrm{a}</em>{1}=\mathbf{1 0} \frac{1}{2}-7=\frac{21}{2}-7=\frac{7}{2}

If this A.P.’s term is 84, then, according to the term formula,

a_{n}=a(n-1) d

84=7+(n-1) \times 7 / 2

77=(n-1) \times 7 / 2

22=n-1

n=23

We all know that, the sum of the \mathrm{n} term is;

\mathrm{S}<em>{\mathrm{n}}=\mathrm{n} / 2(\mathrm{a}+\mathrm{l}), \mathrm{l}=84 \mathrm{S}</em>{\mathrm{n}}=23 / 2(7+84)

\mathrm{S}_{\mathrm{n}}=(23 \times 91 / 2)=2093 / 2

S_{n}=1046 \frac{1}{2}

(ii) Provided that, 34+32+30+\ldots \ldots \ldots \ldots+10

For the given A.P.,

The first term, a=34

The common difference, d=a_{2}-a_{1}=32-34=-2

n^{\text {th }} term, a_{n}=10

Assume that the term of this A.P. is 10,

a_{n}=a+(n-1) d

10=34+(n-1)(-2)

-24=(n-1)(-2)

12=n-1

n=13

We all know that, the sum of the \mathrm{n} terms is;

S_{n}=n / 2(a+l), l=10

=13 / 2(34+10)

=(13 \times 44 / 2)=13 \times 22

=286

i

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