Question

how many three digit number are divisible by 7? find their sum?​

Answer 1

Answer:

Step-by-step explanation:

(i) The 3 - digit numbers divisible by 7 are as follows ;

105, 112, 119, ......, 994.

Clearly, these numbers form an AP with,

a = 105

d = (112 - 105) = 7

last term = 994

Let the total number of terms be n. Then,

Tₙ = 994

⇒ a + (n - 1)d = 994

⇒ 105 + (n - 1) * 7 = 994

⇒ 105 + 7n - 7 = 994

⇒ 7n + 98 = 994

⇒ 7n = 994 - 98

⇒ 7n = 896

⇒ n =  

⇒ n = 128

Hence, there are 128 three-digit numbers divisible by 7.

(ii) Find the sum of terms of the AP : 4, 9, 14,....., 89.

Here,

a = 4

d = (9 - 4) = 5

l = 89

Let the total number of terms be n. Then,

Tₙ = 89

⇒ a + (n - 1)d = 89

⇒ 4 + (n - 1) * 5 = 89

⇒ 4 + 5n - 5 = 89

⇒ 5n - 1 = 89

⇒ 5n = 89 + 1

⇒ 5n = 90

⇒ n =  90/5

⇒ n = 18

Required sum =  n/2 .(a+1)

⇒ Sum =  18/2 .(4+89)

⇒ Sum = 9 * 93

⇒ Sum = 837

Hence, the required sum is 837.

Answer 2

plz refer to this attachment