Question

plzz answer ques. no. 15 n 16..

Answer 1

Answer of 15th question is 14

Answer 2

15)Solution :-
Here,
AP: 105, 112, 119, ........ ,189,196
Now, here we have,

a = 108
d = 7
an = 196
so,
according to the AP formula, we have to find no. of terms that is n.
Now,
an = a + (n-1)d
196 = 105 + (n-1)d
196 = 105 + (n-1)7
196 = 105 + 7n - 7
196 = 98 + 7n
98 = 7n
7n = 98
n = 98/7
n = 14
so, there are 14 term in the given AP.
Now,
by using sum if the term formula, we have
Sn = n[2a + (n - 1)d]/2
Sum of the 14th term = 14[2(105) + (14-1)7] / 2
= 14[210 + (13)7] / 2
= 7[210 + 91]
= 7(301)
= 2107
alternative method :-
Sn = n(a + l) / 2.
[where a = first term and l = last term]
so,
Sn = 14(105+196)/2
Sn = 14(301)/ 2
Sn = 4214 / 2
Sn = 2107

Hence the sum of integer between 100 and 200 which is divisible by 7 is 2107.