6 and 7 both

Q6. If  3   + 5 + 7   + . . . . upto   n   terms   5 + 8 + 11   + . . . . upto   10   terms = 7   ,   then   the   value   of   n   is

        (A) 35                  (B) 36                 (C) 37                  (D) 40


Q7. The sum of integers from 1 to 100 which are divisible by 2 or 5 is :
     
          (A) 300              (B) 3050             (C) 3200              (D) 3250
 

(7).
 

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

⇒ 5n = 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 = 10n

⇒ 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.

Kindly ask the remaining query in different thread.

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