Arithmetic Mean in AP - Practice Questions & MCQ
Quick Facts
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Arithmetic Mean is considered one the most difficult concept.
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25 Questions around this concept.
Solve by difficulty
EASYIf are in A.P. then
equals:
Let G be the geometric mean of two positive numbers a and b, and M be the arithmetic mean of and
if is 4:5 then a:b can be:
For any three positive real numbers a, b and c, 9(25a2+b2)+25(c2−3ac)=15b(3a+c). Then:
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The number of ways in which 5 boys and 3 girls can be seated on a round table if a particular boy $\mathrm{B}_1$ and a particular girl $\mathrm{G}_1$ never sit adjacent to each other, is :
Mean of 100 observations is 45. It was later found that two observations 19 and 31 were incorrectly recorded as 91 and 13. The correct mean is:
N AMs are inserted between 2 and 38. If the third AM is 14, then n is equal to
The average of n numbers $x_1, x_2, x_3, \ldots \ldots x_n$ is $M$. If $x_n$ is replaced by $x^1$ then new average is
5 GMs and 5 AMs are inserted between 1 and 64. If P is the product of these 5 GMs and S is the sum of these 5 AMs, then P + S equals
For some $\mathrm{n} \neq 10$, let the coefficients of the $5^{\text {th }}, 6^{\text {th }}$ and $7^{\text {th }}$ terms in the binomial expansion of $(1+x)^{n+4}$ be in A.P. Then the largest coefficient in the expansion of $(1+x)^{n+4}$ is:
Concepts Covered - 2
Arithmetic Mean
If three terms are in $A P$, then the middle term is called the Arithmetic Mean (A.M.) of the other two numbers. So if a, b and c are in A.P., then $b$ is $A M$ of a and $c$. If $a_1, a_2, a_3, \ldots \ldots, a_n$ are $n$ positive numbers, then the Arithmetic Mean of these numbers is given by
$
A=\frac{a_1+a_2+a_3+\ldots \ldots+a_n}{n}
$
Insertion of $\mathbf{n}$-Arithmetic Mean Between $\mathbf{a}$ and $\mathbf{b}$
If $\mathrm{A}_1, \mathrm{~A}_2, \mathrm{~A}_3 \ldots, \mathrm{~A}_{\mathrm{n}}$ are n arithmetic mean between two numbers a and b, then, $a, \mathrm{~A}_1, \mathrm{~A}_2, \mathrm{~A}_3 \ldots, \mathrm{~A}_{\mathrm{n}}, b$ is an $\mathrm{A}. \mathrm{P}$.
Let d be the common difference of this A.P. Clearly, this A.P. contains $\mathrm{n}+2$ terms.
$
\begin{aligned}
& \therefore b=(n+2)^{t h} \text { term } \\
& \Rightarrow b=a+((n+2)-1) d=a+(n+1) d \\
& \Rightarrow d=\frac{b-a}{n+1}
\end{aligned}
$
Now,
$
\begin{aligned}
& A_1=a+d=\left(a+\frac{b-a}{n+1}\right) \\
& A_2=a+2 d=\left(a+\frac{2(b-a)}{n+1}\right) \\
& \ldots \\
& \cdots \\
& A_n=a+n d=\left(a+n\left(\frac{b-a}{n+1}\right)\right)
\end{aligned}
$
The sum of the n arithmetic mean between two numbers is n times the single A.M. between them.
Proof:
Let $\mathrm{A}_1, \mathrm{~A}_2, \mathrm{~A}_3 \ldots, \mathrm{~A}_{\mathrm{n}}$ be n arithmetic mean of two numbers a and b.
Then, $a, \mathrm{~A}_1, \mathrm{~A}_2, \mathrm{~A}_3 \ldots, \mathrm{~A}_{\mathrm{n}}, b$ is an A.P. with common difference $\frac{b-a}{n+1}$.
Now,
$
\begin{aligned}
& \mathrm{A}_1+\mathrm{A}_2+\mathrm{A}_3+\ldots \ldots+\mathrm{A}_{\mathrm{n}} \\
& =\frac{\mathrm{n}}{2}\left[\mathrm{~A}_1+\mathrm{A}_{\mathrm{n}}\right] \\
& =\frac{\mathrm{n}}{2}[a+b] \\
& {\left[\because a, \mathrm{~A}_1, \mathrm{~A}_2, \ldots, \mathrm{~A}_{\mathrm{n}}, \mathrm{~b} \text { is an A.P., } \therefore a+b=A_1+A_n\right]} \\
& =\mathrm{n}\left(\frac{a+b}{2}\right) \\
& =\mathrm{n} \times(\text { A.M. between } a \text { and } b)
\end{aligned}
$
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