Question

Suppose p is the first of n(n>1) Am’s between two positive numbers a and b, then the value of p is
A. $\dfrac{{na + b}}{{n + 1}}$
B. $\dfrac{{na - b}}{{n + 1}}$
C. $\dfrac{{na + a}}{{n + 1}}$
D. $\dfrac{{na - a}}{{n + 1}}$

Answer 1

In the question given, first we have to find the common difference of the arithmetic progression, when n arithmetic means are inserted between two numbers a and b, this is can be done by using the formula for the $n^{th}$ term in A.P i.,e.,${T_n} = a + \left( {n - 1} \right)d$, where a is the first term, d is the common difference and n is the term, and finally we will find the value of p.

Complete step-by-step answer:
Arithmetic progression or AP is a series of numbers in which the difference between any consecutive terms will be the same.
Given that p is the first of n(n>1) Am’s between two positive numbers a and b,
Now let us consider,
$A{M_1},A{M_2},A{M_3}......A{M_n}$ be n AM’s between a and b,
So then $a,A{M_1},A{M_2},A{M_3}......A{M_n},b$ are in Arithmetic progression.
Now from the given data, $p = A{M_1}$,
The series becomes$a,p,A{M_2},A{M_3}......A{M_n},b$,
Now $p,A{M_2},A{M_3}......A{M_n}$ are in AP according to the given question.
Now from then above $b = {\left( {n + 2} \right)}$ term,
Now using $n^{th}$ formula i.e.,
${T_n} = a + \left( {n - 1} \right)d$, where a is the first term, d is the common difference, n is the nth term and
Here, ${T_n} = b,a = a,n = n + 2,d = d$, now substituting we get,
$\Rightarrow$ $b = a + \left( {\left( {n + 2} \right) - 1} \right)d$,
Now simplifying we get,
$\Rightarrow$ $b = a + \left( {n + 1} \right)d$,
And taking a to the L.H.S we get,
$\Rightarrow$ $b - a = \left( {n + 1} \right)d$,
Now taking n+1 term to L.H.S we get,
$\Rightarrow$ $d = \dfrac{{b - a}}{{n + 1}} - - - - (1)$,
Now we also should find $p^{th}$ term so we have,
$\Rightarrow$ $p = a + (2 - 1)d$, as p is the second term then,
$\Rightarrow$ $p = a + d$,
Now taking to L.H.S we get,
$\Rightarrow$ $d = p - a - - - - (2)$,
So equate the (1) and (2) we get,
$\Rightarrow$ $\dfrac{{b - a}}{{n + 1}} = p - a$,
Now simplifying we get,
$\Rightarrow$ $p = \dfrac{{b - a}}{{n + 1}} + a$,
Now taking L.C.M we get,
$\Rightarrow$ $p = \dfrac{{b - a + a\left( {n + 1} \right)}}{{n + 1}}$,
Now multiplying we get,
$\Rightarrow$ $p = \dfrac{{b - a + an + a}}{{n + 1}}$,
Now eliminating the terms we get,
$\Rightarrow$ $p = \dfrac{{b + an}}{{n + 1}}$.
So the value of$p = \dfrac{{b + an}}{{n + 1}}$,

If p is the first of n(n>1) Am’s between two positive numbers a and b, then the value of p is $\dfrac{{b + an}}{{n + 1}}$.

Note:
IIn these type of questions, students should have clear idea about the progression in which the given series is in and what should be found and which formula should be used as there are three progressions i.e., arithmetic , geometric and harmonic progressions and can also make mistakes in which formula should be used in the question.