Question

How many term of two digit are divisible by 7​

Answer 1

by 7 = 14

last two digit number divisible by 7 = 98

so a = 14 and l = 98 where a = first term of A.P. and l = last term of A.P.

so our A.P. formed is 14 , ….. , …., .. , 98

formula for finding the no of terms in A.P =

l = a + (n-1) d

(d means difference between terms = 7)

so now put the values

i.e. 98 = 14 + (n - 1) 7

84 = 7n - 7

91 = 7n

n = 91/7 = 13

Answer 2

two digit no. divisible by 7:-

14,21,28,35__________98

therefore a=14 and d=21–14=7 and l=98

l=a+(n-1)d

98=14+(n-1)7

98–14=7n-7

84=7n-7

84+7=7n

91=7n

91÷7=n

13=n