FIND THE SUM OF ALL MULTIPLES BETWEEN TWO NUMBERS

Question 1 :

How many three digit numbers are divisible by 7?

Solution :

The first three digit number is 100 and the last three digit number is 999

Now we should find the number of three digit terms which are divisible by 7.

105 is the first three digit number divisible by 7 and 994 is the last three digit number which is divisible by 7.

 105 + 112 + ........... + 994

an = a + (n - 1) d

a = 105, d = 112 - 105  = 7 

994  =  105 + (n - 1)7

994 - 105  =  (n - 1)7

889  =  (n- 1)7

889/7  =  n - 1

127  =  n - 1

n  =  127 + 1

n  =  128

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

Question 2 :

How many multiples of 4 lie between 10 and 250? 

Solution :

12, 16, 20,.......................,248

 an = a + (n - 1) d

a = 12      d = 16 - 12 = 4       an = 248   

248  =  12 + (n - 1) 4

248 - 12  =  (n - 1) 4

236  =  (n - 1) 4

236/4  =  n - 1

59  =  n - 1

n  =  59 + 1

n  =  60

Question 3 :

For what value of n, are the nth terms of two APs 63, 65, 67,....... and 3, 10, 17,........ equal?

Solution :

Let an and b n are two nth terms of the given sequence respectively.

an = a + (n - 1) d

a = 63, d = 65 - 63 = 2

an = 63 + (n - 1) 2 ---(1)

bn = b + (n - 1) d

b = 3, d  =  10 - 3 = 7

bn = 3 + (n - 1) 7---(2)

(1)  =  (2)

an = bn

63 + (n - 1) 2  =  3 + (n - 1) 7

63 + 2 n - 2  =  3 + 7n - 7

61 + 4  =  7 n - 2 n

65  =  5 n

n  =  65/5

n  =  13  

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