Question

Find 14 A. M's between 5 and 8 and hence show that their sum is 14 times the AM.
between 5 and 8.
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Answer 1

Step-by-step explanation:

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

for that

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

Step1: Finding AP

Here a= 5

b=8

n= 14

$d = \frac{8 - 5}{14 + 1} \\ \\ d = \frac{1}{5}$

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

$S_n = \frac{n}{2} (a + l) \\ \\ a = \frac{26}{5} \\ \\ l = \frac{39}{5} \\ \\ n = 14\\ \\ S_{14}= \frac{14}{2} ( \frac{26}{5} + \frac{39}{5} ) \\ \\ = 7( \frac{65}{5} ) \\ \\ =7 \times 13 \\ \\ S_{14}= 91 ...eq1$

Step 3: Find AM

$AM= \frac{5 + 8}{2} \\ \\ = 6.5$

Step 4: Find 14 times of AM

$= 14 \times 6.5 \\ \\ = 91...eq2$

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

Hope it helps you.

Answer 2

The answer of the answer of this question is = 91