Question

the number of term of ap 1/3,1/2,2/3,.....11/6 is

Answer 1

Step-by-step explanation:

let there is n terms of this ap.

a = first term of this ap = 1/3

d= common difference of this ap = 1/2-1/3= 1/6

so 11/6 is nth term of ap.

so Tn = a+ (n-1)d

11/6= 1/3 + (n-1).1/6

so n =( 9.6)/6 + 1

n= 10

so there are 10 terms of this ap

Answer 2

Number of terms in the AP is 10

Given :

The arithmetic progression 1/3 , 1/2 , 2/3 , . . . . , 11/6

To find :

The number of terms in the arithmetic progression

Concept :

If in an arithmetic progression

First term = a

Common difference = d

Then nth term of the AP

= a + ( n - 1 )d

Solution :

Step 1 of 3 :

Write down the given progression

Here the given arithmetic progression is

1/3 , 1/2 , 2/3 , . . . . , 11/6

Step 2 of 3 :

Write down first term and common difference

The arithmetic progression is

1/3 , 1/2 , 2/3 , . . . . , 11/6

First term = a = 1/3

Common Difference = d = 1/2 - 1/3 = 1/6

Step 3 of 3 :

Find the number of terms

Let number of terms in the AP = n

Then nth term of the AP = 11/6

$\displaystyle \sf{ a + ( n - 1 )d = \frac{11}{6} }$

$\displaystyle \sf{ \implies \frac{1}{3} + \frac{1}{6} ( n - 1 ) = \frac{11}{6} }$

$\displaystyle \sf{ \implies \frac{1}{6} ( n - 1 ) = \frac{11}{6} - \frac{1}{3} }$

$\displaystyle \sf{ \implies \frac{1}{6} ( n - 1 ) = \frac{11 - 2}{6} }$

$\displaystyle \sf{ \implies \frac{1}{6} ( n - 1 ) = \frac{9}{6} }$

$\displaystyle \sf{ \implies ( n - 1 ) = 9 }$

$\displaystyle \sf{ \implies n = 9 + 1 }$

$\displaystyle \sf{ \implies n = 10}$

There are 10 terms in the arithmetic progression

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