Question

There are n a.m's between 3 and 17 ratio of last to first mean is 3:1 .find n

Answer 1

3+d/17-d =1/3
9+3d=17-d
d=2
14/2=n+1
n=6 is the answer

Answer 2

Answer:

Number of terms are n=6.

Step-by-step explanation:

Given : There are n a.m's between 3 and 17 ratio of last to first mean is 3:1.

To find : The value of n ?

Solution :

According to question, The A.P form is

$3,A_1,A_2,A_3,....A_n,17$

Here, a=3 and $a_{n+2}=17$

We know, $a_n=a+(n-1)d$

So, $a+(n+2-1)d=17$    

$3+(n+1)d=17$                            

$(n+1)d=14$        

$d=\frac{14}{n+1}$        

We have given,

$\frac{A_n}{A_1}=\frac{3}{1}$

$\frac{a_{n+1}}{a_2}=\frac{3}{1}$

$\frac{a+(n+1-1)d}{a+d}=\frac{3}{1}$

$\frac{3+(n)d}{3+d}=\frac{3}{1}$

$3+(n)d=9+3d$

Substitute d,

$3+(n)(\frac{14}{n+1})=9+3(\frac{14}{n+1})$

$3+\frac{14n}{n+1}=9+\frac{42}{n+1}$

$\frac{14n}{n+1}-\frac{42}{n+1}=6$

$\frac{14n-42}{n+1}=6$

$14n-42=6n+6$

$14n-6n=42+6$

$8n=48$

$n=\frac{48}{8}$

$n=6$

Therefore, Number of terms are n=6.