find that the sum of all prime numbers from 1 to 100 is divisible by 6
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Natural no less than 100 and are divisible by 4 are 4, 8, 12, 16, …, 96
Clearly this is an AP, with
First term, a = 4
Common difference, d = 4
Last term or nth term, an = 96
We know that, nth term of an AP is,
an = a + (n - 1)d
where a is first term and d is common difference of AP
Putting values, we get
96 = 4 + (n - 1)4
⇒ 92 = 4n - 4
⇒ 96 = 4n
⇒ n = 24
Also, sum of an AP if first and nth term are given,
Where, a is first and an is nth term
JUST PUT 6 INSTEAD OF 4
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