Answer:
867
Step-by-step explanation:
a = 3, d = 6, nth term = 99
nth term = a + (n - 1)(d)
=> 99 = 3 + (n - 1)(6)
=> 3 + 6n - 6 = 99
=> 6n - 3 = 99
=> 6n = 99 + 3 = 102
=> n = 102/6 = 17
∴ Sum = n/2 (a + nth term)
= 17/2 (3 + 99)
= 17/2 * 102 = 17 * 51 = 17 * 17 * 3 = 289 * 3 = 867
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Find the sum of all odd integers between 2 and 100 divisible by 3.
Easy
Solution
verified
Verified by Toppr
The odd integers between 2 and 100 which are divisible by 3 are 3, 9, 15, 21, ..., 99.
Here, a=3 and d=6.
a
n
=99
⇒a+(n−1)d=99
⇒3+(n−1)×6=99
⇒6(n−1)=96
⇒n−1=16
⇒n=17
Therefore,
Required sum =
2
(a+l)=
17
(3+99)=867
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Answers & Comments
Answer:
867
Step-by-step explanation:
a = 3, d = 6, nth term = 99
nth term = a + (n - 1)(d)
=> 99 = 3 + (n - 1)(6)
=> 3 + 6n - 6 = 99
=> 6n - 3 = 99
=> 6n = 99 + 3 = 102
=> n = 102/6 = 17
∴ Sum = n/2 (a + nth term)
= 17/2 (3 + 99)
= 17/2 * 102 = 17 * 51 = 17 * 17 * 3 = 289 * 3 = 867
Answer:
search-icon-header
Search for questions & chapters
search-icon-image
Question
Bookmark
Find the sum of all odd integers between 2 and 100 divisible by 3.
Easy
Solution
verified
Verified by Toppr
The odd integers between 2 and 100 which are divisible by 3 are 3, 9, 15, 21, ..., 99.
Here, a=3 and d=6.
a
n
=99
⇒a+(n−1)d=99
⇒3+(n−1)×6=99
⇒6(n−1)=96
⇒n−1=16
⇒n=17
Therefore,
Required sum =
2
n
(a+l)=
2
17
(3+99)=867