Question

Find 14 A.M's between 5 and 8 and hence show that their sum is 14 times the A.M. between 5 and 8.​

Answer 1

Answer:

We know that ,one can insert n AM between two numbers,so that the given sequence remains an AP.

for that

\begin{gathered}d = \frac{b - a}{n + 1}

Step1: Finding AP

Here a= 5

b=8

n= 14

Now,the AP so formed has first term a,second term a+d,3rd term a+2d and so on

Thus the AP is

5,26/5,27/5,28/5,29/5,30/5... 39/5,8

Step2: Find sum of inserted AMs

26/5,27/5,28/5,29/5,30/5... ,38/5,39/5

(526+539)=7(565)=7×13S14=91...eq1

Step 3: Find AM

\begin{gathered} AM= \frac{5 + 8}{2} \\ \\ = 6.5 \\ \\ \end{gathered}AM=25+8=6.5

Step 4: Find 14 times of AM

\begin{gathered} = 14 \times 6.5 \\ \\ = 91...eq2 \\ \\ \end{gathered}=14×6.5=91...eq2

It is clear from eq1 and eq2,that sum of AMs is 14 times the AM.

Hope it helps you.