Question

If ‘n’ arithmetic means are inserted between 2 and 38. if third arithmetic mean is 14 then n is equal to
A) 9
B) 7
C) 8
D) 10

Answer 1

Here we are given the first and last term of the series. Consider it as arithmetic progression. Also we are given a third arithmetic mean and with this we will find the common difference. Then with the help of the last term formula of A.P. we will find the number of terms.

Complete step by step solution:
Given that n arithmetic means are inserted between 2 and 38.
So let’s assume it is an A.P.
First-term a=2
Last term =38
total number of terms= n+2
the common difference is d.
So A.P.is 2, $a_1, a_2, a_3,.....a_n$, 38.
But the third arithmetic mean is 14.
$a_3 = 14$

$$\Rightarrow 2 + 3d = 14 \\\ \Rightarrow 3d = 12 \\\ \Rightarrow d = 4 \\\$$

Now last term in an A.P. is given by$a + \left[ {n - 1} \right]d$,

$$\Rightarrow 2 + \left[ {n + 2 - 1} \right]4 = 38 \\\ \Rightarrow 2 + \left[ {n + 1} \right]4 = 38 \\\ \Rightarrow 4n + 4 = 36 \\\ \Rightarrow 4n = 32 \\\ \Rightarrow n = 8 \\\$$

Thus option C is correct.

Additional information:

1. An arithmetic progression is a series in which two terms are having a common difference between them.
2. a, a+d, a+2d, ….is an A.P.
3. Last term in an A.P. is given by $a + \left[ {n - 1} \right]d$
4. Sum of first n terms in an A.P. is given by the formula ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$.

Note:
There is one more interesting series related to Arithmetic progression is the Harmonic series. Just for the understanding of students, it’s inversely or reciprocal to AP. 3 terms a,b, c are in harmonic progression if $\dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c}$ are in AP.