Question

1. find the sum of all the two digit odd positive integer .

2.find the sum of all multiples of 9 lying between 300 and 700.

3. find the sum of all the three digit natural numbers which are multiples of 7 .

Answer 1

(1)
Sol:
Two digit odd positive numbers are 11,13,15,17...........99 are in A.P.

 Here a = 11 and d = 2, tn= 99, n = ?

Sum of the n terms = (n/2)[2a+(n -1)d]

But tn = a + (n -1)d

⇒ 99 = 11+ (n-1)2

⇒ 99 -11 = (n-1)2

⇒ 88/2 = (n-1)

∴ n = 45.

subsitute n = 45  in sum of the n terms we obtain

⇒ s45 = (45/2)(2×11 + (45 -1)2)

⇒ s45 = (45/2)(110)

⇒ s45 = 45×55.

⇒  s45 = 2475.

∴sum of all two digit odd positive numbers = 2475.



2) See the solution of second question in image

(3)
The first 3-digit multiple of 7 is 7*15 = 105, and the last one is 7*142 = 994. So if you think of it this way: 

7*15 + 7*16 + 7*17 + ...... + 7*142 
= 7 (15+16+17+...+142) 
Then we can find the sum of the arithmetic series in the parentheses and then multiply at the end by 7. 

In this arithmetic series, a(1) = 15, d=1 and n = 142-15+1 = 128. 
Therefore, S(128) = 128/2 (15+142) = 64*157 = 10048. 

Multiply by 7 to get 70336. 

Note: Of course the original series 105 + 112 + 119 + .... + 994 is itself arithmetic with a(1) = 105, d=7 and n = 128.