Question 5

Find the sum of integers from 1 to 100 that are divisible by 2 or 5.

Answer

The integers from 1 to 100, which are divisible by 2, are 2, 4, 6… 100.

This forms an A.P. with both the first term and common difference equal to 2.

⇒100 = 2 + (*n* –1) 2

⇒ *n* = 50

The integers from 1 to 100, which are divisible by 5, are 5, 10… 100.

This forms an A.P. with both the first term and common difference equal to 5.

∴100 = 5 + (*n* –1) 5

⇒ 5*n* = 100

⇒ *n* = 20

The integers, which are divisible by both 2 and 5, are 10, 20, … 100.

This also forms an A.P. with both the first term and common difference equal to 10.

∴100 = 10 + (*n* –1) (10)

⇒ 100 = 10*n*

⇒ *n* = 10

∴Required sum = 2550 + 1050 – 550 = 3050

Thus, the sum of the integers from 1 to 100, which are divisible by 2 or 5, is 3050.

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The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.

Priyanshu
2019-11-13 14:16:36

Thanks for the solution.

Shivanshu
2019-10-16 20:38:56

That is correct but I have asked other answer

Aarya
2019-09-16 17:47:16

Great..

Deekshitha
2019-09-16 05:44:54

Thanks sir it helped me

jitendra pushkar
2019-09-13 15:50:28

thanks sir.. ... wonderful answer.....

Subhash
2019-09-02 14:27:38

Thanks a lot

Ashutosh kumar
2019-08-26 16:05:50

Thanks sir

Asha
2019-08-23 16:04:41

Can it be solved by n(n+2)/2

Gunjan
2019-06-04 17:57:04

Thanks

Vivek singh
2019-05-18 21:14:49

Not best

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