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Question

A. Find the sum of the following arithmetic series.

(1) 2 + 5 + 8 + ……. upto 12 terms

(2) 1 + 4 + 7 + ……..upto 22 terms

(3) 3 + 7 + 11 + …….upto 35 terms

(4) Find the sum of arithmetic series which contains 25 terms and whose middle term is 20.

B. In an A.P. if

(1) Tn = 4n + 3 find S15

(2) Tn = 5 − 2n find S20

C. (1) Find the sum of all the first ‘n’ odd Natural numbers.

(2) Find the sum of all Natural numbers between 101 and 201 which are divisible by 4.

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Solution

(A) The sum of the first n terms of an A.P. is,

(1)

2, 5, 8 ... forms an A.P. with and

The sum of the given series up to 12 terms can be calculated as:

(2)

1, 4, 7 … forms an A.P. with and

The sum of the given series up to 22 terms can be calculated as:

(3)

3, 7, 11 … forms an A.P. with and

The sum of the given series up to 35 terms can be calculated as:

(4) It is given that n = 25 and the middle term is 20 i.e., thirteenth term, = 20.

Hence, the sum of the arithmetic series is 500.

(B)

(1) Tn = 4n + 3

(2) Tn = 5 2n

(C)

(1) The first n odd natural numbers are 1, 3, 5, 7, 9 … n

It can be observed that these numbers form an A.P. whose first term is 1 and common difference is 2 i.e., and .

Hence, the sum of all the first n odd natural numbers is .

(2) The list of all natural numbers between 101 and 201, which are divisible by 4 is:

104, 108 … 200

It can be observed that these numbers form an A.P. whose first term is 104, last term is 200 and common difference is 4 i.e., and.


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