Question

There are n A. M. s between 5 and 35 such that 2nd mean : last mean 1:4 find n

Answer 1

Answer:

The value of n is 17

Step-by-step explanation:

$\text{Let }A_1,A_2,.....A_n\text{ be n arithmetic means between 5 and 35}$

$\text{Then,}$

$A_1=5+d$

$A_2=5+2d$

....

.....

$A_n=5+nd$

$35=5+(n+1)d$

$\implies\,30=(n+1)d$........(1)

$\text{Also,}$

$A_2:A_n=1:4$

$\implies\frac{A_2}{A_n}=\frac{1}{4}$

$\implies\frac{5+2d}{5+nd}=\frac{1}{4}$

$\implies\,20+8d=5+nd$

$\implies\,15=(n-8)d$

$\implies\,d=\frac{15}{n-8}$..........(2)

$\text{using (2) in (1), we get }$

$30=(n+1)\frac{15}{n-8}$

$\implies\,2=\frac{n+1}{n-8}$

$\implies\,2n-16=n+1$

$\implies\,\boxed{\bf\,n=17}$

Answer 2

The value of n is 17.

Step-by-step explanation:

We are given that there are n A. M. s between 5 and 35 such that 2nd mean: last mean 1:4.

Let the series be 5, $A_1, A_2,A_3, ..........,A_n$, 35 which means there are n A.M.'s between 5 and 35.

Here a = first term = 5 and let d = common difference.

So, 2nd mean = 3rd term of the series

$A_3$ = a + (n - 1)d = a + 2d

Simiarly, last mean = (n+1)th term of the series

        $A_n_+_1$ = a + (n - 1)d = a + (n + 1 - 1)d = a + nd  

Now, the ratio given to us is;

                   $\frac{a+2d}{a+nd}=\frac{1}{4}$

                 4a + 8d = a + nd

                   nd = 3a + 8d

                   nd = 15 + 8d       {as given a = 5} ------------ [Equation 1]

Now, as we know that if there are n arithmetic means between two number, then there are (n+2) terms in an A.P.

So, $A_n_+_2$ = a + (n - 1)d = a + (n + 2 - 1)d = a + (n + 1)d  

           35  =  5 + (n + 1)d

           30 = nd + d

           nd = 30 - d

Now, putting value of nd in equation 1 we get;

             30 - d = 15 + 8d

             8d + d = 30 - 15

                  d =  $\frac{15}{9}$

Put this value of d in equation 1 we get;

                   nd = 15 + 8d

                     n  =  $\frac{15+8d}{d}$

                     n  =  $\frac{15+(8\times \frac{15}{9}) }{\frac{15}{9} }$

                     n  =  $\frac{135 + 120}{9} \times \frac{9}{15}$  = $\frac{255}{15}$

                     n  = 17