If six arithmetic means are inserted between 3 and 13/2, the ratio of the fifth and the first mean is: Option 1: 3 : 2 Option 2: 11 : 7 Option 3: 9 : 8 Option 4: 8 : 5

Answer: 11 : 7. Solution: When we insert 6 arithmetic means between 3 and 13/2, we create an arithmetic progression with 8 terms total. - First term: a_1 = 3 - Last term: a_8 = 13/2 - The 6 arithmetic means are: a_2, a_3, a_4, a_5, a_6, a_7 In an AP: a_n = a_1 + (n-1)d So: a_8 = a_1 + 7d 13/2 = 3 + 7d 13/2 - 3 = 7d 13/2 - 6/2 = 7d 7/2 = 7d d = 1/2 First arithmetic mean: a_2 = 3 + (1)(1/2) = 7/2 Fifth arithmetic mean: a_6 = 3 + (5)(1/2) = 3 + 5/2 = 11/2 Ratio = Fifth mean : First mean = 11/2 : 7/2 = 11 : 7 Answer: 11:7

Algebra

Progression & Series

Medium

If six arithmetic means are inserted between 3 and $\frac{13}{2}$ , the ratio of the fifth and the first mean is:

3 : 2

11 : 7

9 : 8

8 : 5

✅ Correct Option: 2

When we insert $6$ arithmetic means between $3$ and $\dfrac{13}{2}$ , we create an arithmetic progression with $8$ terms total.

First term: $a_1 = 3$
Last term: $a_8 = \dfrac{13}{2}$
The $6$ arithmetic means are: $a_2, a_3, a_4, a_5, a_6, a_7$

In an AP: $a_n = a_1 + (n-1)d$

So: $a_8 = a_1 + 7d$

$\dfrac{13}{2} = 3 + 7d$

$\dfrac{13}{2} - 3 = 7d$

$\dfrac{13}{2} - \dfrac{6}{2} = 7d$

$\dfrac{7}{2} = 7d$

$d = \dfrac{1}{2}$

First arithmetic mean: $a_2 = 3 + (1)\left(\dfrac{1}{2}\right) = \dfrac{7}{2}$

Fifth arithmetic mean: $a_6 = 3 + (5)\left(\dfrac{1}{2}\right) = 3 + \dfrac{5}{2} = \dfrac{11}{2}$

Ratio = Fifth mean : First mean = $\dfrac{11}{2} : \dfrac{7}{2} = 11 : 7$

Answer: $11:7$

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If six arithmetic means are inserted between 3 and $\frac{13}{2}$ , the ratio of the fifth and the first mean is: