if four numbers are in ap such that their sum is 32 and the least numb

1.	if four numbers are in ap such that their sum is 32 and the least number is one seventh the greatest number, then find the four numbers
Answer» $\large\underline{\sf{Given- }}$ Four numbers are in AP whom sum is 32 Least NUMBER is ONE seventh of greatest number. $\large\underline{\sf{To\:Find - }}$ Four numbers of AP. $\large\underline{\bold{Solution-}}$ Since, four numbers are in AP, Therefore, $\begin{gathered}\begin{gathered}\bf \:Let \: numbers \: be - \begin{cases} &\sf{a - 3d} \\ &\sf{a - d}\\ &\sf{a + d}\\ &\sf{a + 3d} \end{cases}\end{gathered}\end{gathered}$ According to statement, Sum of 4 numbers is 32 $\rm :\implies\:a - \cancel{3d} + a - \cancel{d} + a + \cancel{d} + a + \cancel{3d} = 32$ $\rm :\longmapsto\:4a = 32$ $\bf\implies \:a = 8 - - - (1)$ Now, Again, it is given that, least number is one seventh of greatest. $\rm :\longmapsto\:a - 3d = \dfrac{1}{7} (a + 3d)$ $\rm :\longmapsto\:7a - 21d = a + 3d$ $\rm :\longmapsto\:6a = 24d$ $\rm :\longmapsto\:a = 4d$ $\rm :\implies\:4d = 8 \: \: \: \: (\because \: a = 8)$ $\bf\implies \:d = 2 - - (2)$ On substituting the values of a and d, we have $\begin{gathered}\begin{gathered}\bf \:Hence, \: numbers \: are - \begin{cases} &\sf{a - 3d = 8 - 6 = 2} \\ &\sf{a - d = 8 - 2 = 6}\\ &\sf{a + d = 8 + 2 = 10}\\ &\sf{a + 3d = 8 + 6 = 14} \end{cases}\end{gathered}\end{gathered}$ Additional INFORMATION :- ↝ nᵗʰ term of an arithmetic sequence is, $\begin{gathered}\:{\underline{{\boxed{\bf{{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \end{gathered}$ Wʜᴇʀᴇ, aₙ is the nᵗʰ term. a is the FIRST term of the sequence. n is the no. of terms. d is the common difference. ↝ Sum of first 'n' terms of an arithmetic sequence is, $\begin{gathered}\:{\underline{{\boxed{\bf{{S_{n}\:=\dfrac{n}{2} (2\:a\:+\:(n\:-\:1)\:d)}}}}}} \end{gathered}$ Wʜᴇʀᴇ, Sₙ is the sum of first n terms a is the first term of the sequence. n is the no. of terms. d is the common difference.

if four numbers are in ap such that their sum is 32 and the least number is one seventh the greatest number, then find the four numbers

Answer»

$\large\underline{\sf{Given- }}$

Four numbers are in AP whom sum is 32

Least NUMBER is ONE seventh of greatest number.

$\large\underline{\sf{To\:Find - }}$

Four numbers of AP.

$\large\underline{\bold{Solution-}}$

Since, four numbers are in AP,

Therefore,

$\begin{gathered}\begin{gathered}\bf \:Let \: numbers \: be - \begin{cases} &\sf{a - 3d} \\ &\sf{a - d}\\ &\sf{a + d}\\ &\sf{a + 3d} \end{cases}\end{gathered}\end{gathered}$

According to statement,

Sum of 4 numbers is 32

$\rm :\implies\:a - \cancel{3d} + a - \cancel{d} + a + \cancel{d} + a + \cancel{3d} = 32$

$\rm :\longmapsto\:4a = 32$

$\bf\implies \:a = 8 - - - (1)$

Now,

Again,

it is given that,

least number is one seventh of greatest.

$\rm :\longmapsto\:a - 3d = \dfrac{1}{7} (a + 3d)$

$\rm :\longmapsto\:7a - 21d = a + 3d$

$\rm :\longmapsto\:6a = 24d$

$\rm :\longmapsto\:a = 4d$

$\rm :\implies\:4d = 8 \: \: \: \: (\because \: a = 8)$

$\bf\implies \:d = 2 - - (2)$

On substituting the values of a and d, we have

$\begin{gathered}\begin{gathered}\bf \:Hence, \: numbers \: are - \begin{cases} &\sf{a - 3d = 8 - 6 = 2} \\ &\sf{a - d = 8 - 2 = 6}\\ &\sf{a + d = 8 + 2 = 10}\\ &\sf{a + 3d = 8 + 6 = 14} \end{cases}\end{gathered}\end{gathered}$

Additional INFORMATION :-

↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\:{\underline{{\boxed{\bf{{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \end{gathered}$

Wʜᴇʀᴇ,

aₙ is the nᵗʰ term.

a is the FIRST term of the sequence.

n is the no. of terms.

d is the common difference.

↝ Sum of first 'n' terms of an arithmetic sequence is,

$\begin{gathered}\:{\underline{{\boxed{\bf{{S_{n}\:=\dfrac{n}{2} (2\:a\:+\:(n\:-\:1)\:d)}}}}}} \end{gathered}$

Wʜᴇʀᴇ,

Sₙ is the sum of first n terms

a is the first term of the sequence.

n is the no. of terms.

d is the common difference.