1.

II. There are 9 multiples of 13 from 19 to 119.1). Only I2). Only II3). Both I and II4). Neither I nor II

Answer»

Considering each statement one by one:

Statement I:

Multiples of 9 between 7 and 109 are 9, 18, 27, …., 108

⇒ The above series is an AP with FIRST term = 9 and common difference = 9 and last term = 108

In an AP, An = a + (n - 1) × d

$(\Rightarrow 108 = 9 + \left( {n - 1} \right)9 = 9 + 9n - 9)$

⇒ 108 = 9n

⇒ n = 108/9 = 12

∴ Statement I is correct.

Statement II:

Multiples of 13 from 19 to 119 are 26, 39, …., 117

The above series is an AP with first term = 26 and common difference = 13 and last term = 117

In an AP, An = a + (n - 1) × d

⇒ $(117 = 26 + \left( {n - 1} \right)13 = 26 + 13n - 13)$

⇒ 117 - 13 = 13n

⇒ 13n = 104

⇒ n = 104/13 = 8

∴ Statement II is incorrect.


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