Question

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 1

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

• 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.