Question

ONLY FOR MASTERS- solve AP​

Answer 1

Question :--- Find rhe number of natural numbers between 95 and 985 which are divisible by 2 and 5 both .

Concept used :--

→ Any number divisible by 2 and 5 both, that means It will be Divisible by 2*5 = 10 also . , or we can say that, if any number divisible by 10 , than only it will be divisible by 2 and 5 both .

→ nth term of AP series is given by :- a + (n-1)d, where , a is first term, and d is common difference ..

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Solution :---

In Question, we have to Find all numbers between 95 and 985 that are divisible by 10 .

So, Our First number will be = a = 100

→ second Number = 110

→ Third Number = 120 ,

so, on

→ Last number = nth number = 980 [ upto 985 we have to Find] .

so, we can say that, our AP series is :-- 100, 110, 120, ___________ 980 .

where ,

→ a = 100

→ d = 120-110 = 10

→ nth term = 980 ,

we have to Find value of n ?

Putting values in above told formula now , we get,

→ 980 = 100 + (n-1)*10

→ 980-100 = 10n - 10

→ 880 + 10 = 10n

→ 890 = 10n

Dividing both sides by 10 now we get,

→ n = 89

Hence, we can say that, Total natural Numbers b/w , 95 and 985 that are divisible by 2 and 5 both are Total 89 .

Answer 2

Answer:

Hello friend your answer is 89

Step-by-step explanation:

The AP formed by the numbers divisible by 2 and 5 is: 100,110,120,130..........980. Now here a= 100 d= a2- a1 d= 110-100 = 10 an= 980 We know that an= a+(n-1)d. 980= 100+(n-1)10. 980=100+10n- 10. 880= 10n- 10. 880+10 = 10n. 890= 10n. n= 89 So 89 natural numbers between 95 and 985 are divisible by 2 & 5 both between