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

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

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

Answer»

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.