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6 and 7 both

Q6. If $\frac{3+5+7+....\mathrm{upto}\mathrm{n}\mathrm{terms}}{5+8+11+....\mathrm{upto}10\mathrm{terms}}=7,\mathrm{then}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{n}\mathrm{is}$

(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

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.

Kindly ask the remaining query in different thread.

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