Question

How many terms are there in the sequence of 3/4,1,5/4,...............3?​

Answer 1

Answer:

Step-by-step explanation:

a₂ - a₁ = 1- 3/4 = 1/4

a₃ - a₂ = 5/4 -1 = 1/4.

So the series is in AP with a = 3/4, d = 1/4, Tₙ = 3

nth term of AP is given by Tₙ = a + (n-1)d

       3 = 3/4 + (n-1)1/4

      (n - 1)1/4 = 3 - 3/4

      (n - 1)1/4   = 9/4

       n - 1 = 9

       n = 10.

Thus there are 10 terms in the series.

Answer 2

Answer :

10 terms

Step-by-step explanation :

$\underline{\bf Arithmetic \ Progression:}$

• It is the sequence of numbers such that the difference between any two successive numbers is constant.

• General form of AP,

      a , a+d , a+2d , a+3d , ..........

____________________________

Given series is of A.P, (since the difference between successive terms is constant)

3/4 , 1 , 5/4 , ........ , 3

• first term,

      a = 3/4

• common difference,

     d = 1 - 3/4 = 5/4 - 1 = 1/4

Let the nth term be 3 (since it is the last term)

   aₙ = 3

we have to find the value of "n"

we know,

 nth term of A.P. is given by,

  $\boxed{\bf a_n=a+(n-1)d}$

  $3=\frac{3}{4} +(n-1)(\frac{1}{4}) \\\\ 3-\frac{3}{4} =(n-1)(\frac{1}{4}) \\\\ \frac{3 \times 4}{4} -\frac{3}{4} =(n-1)(\frac{1}{4}) \\\\ \frac{12}{4} -\frac{3}{4} = (n-1)(\frac{1}{4}) \\\\ \frac{12-3}{4} =(n-1)(\frac{1}{4}) \\\\ \frac{9}{4}=(n-1)(\frac{1}{4}) \\\\ 9=(n-1)(1) \\\\ n-1=9 \\\\ n=9+1 \\\\ n = 9+1 \\\\ \bf n=10$

There are 10 terms in the given sequence of A.P.