Four numbers are in arithmetic progression. T | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(d) Let ${A_1},{A_2},{A_3}$ and ${A_4}$ are four numbers in $A.P.$

${A_1} + {A_4} = 8$ ..$(i)$

and ${A_2}.\,{A_3} = 15$ ..$(ii)$

The sum of t

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(d) Let ${A_1},{A_2},{A_3}$ and ${A_4}$ are four numbers in $A.P.$

${A_1} + {A_4} = 8$ ..$(i)$

and ${A_2}.\,{A_3} = 15$ ..$(ii)$

The sum of terms equidistant from the beginning and end is constant and is equal to

sum of first and last terms.

Hence, ${A_2} + {A_3} = {A_1} + {A_4} = 8$ ..$(iii)$

From $(ii)$ and $(iii),$

${A_2} + \frac{{15}}{{{A_2}}} = 8$

==> $A_2^2 - 8{A_2} + 15 = 0$

${A_2} = 3\,\,{m{or}}\,\,5$ and ${A_3} = 5\,\,\,{m{or}}\,\,{m{3}}$.

As we know, ${A_2} = \frac{{{A_1} + {A_3}}}{2}$

==> ${A_1} = 2{A_2} - {A_3}$

==> ${A_1} = 2 	imes 3 - 5 = 1$ and ${A_4} = 8 - {A_1} = 7$

Hence the series is, $1, 3, 5, 7.$

So that least number of series is $1.$

Four numbers are in arithmetic progression. The sum of first and last term is $8$ and the product of both middle terms is $15$. The least number of the series is

Similar Questions

If the roots of the equation ${x^3} - 12{x^2} + 39x - 28 = 0$ are in $A.P.$, then their common difference will be

If $\log _{3} 2, \log _{3}\left(2^{x}-5\right), \log _{3}\left(2^{x}-\frac{7}{2}\right)$ are in an arithmetic progression, then the value of $x$ is equal to $.....$

If $A$ be an arithmetic mean between two numbers and $S$ be the sum of $n$ arithmetic means between the same numbers, then

The number of common terms in the progressions $4,9,14,19, \ldots \ldots$, up to $25^{\text {th }}$ term and $3,6,9,12$, up to $37^{\text {th }}$ term is :

Find the $17^{\text {th }}$ and $24^{\text {th }}$ term in the following sequence whose $n^{\text {th }}$ term is $a_{n}=4 n-3$