Question

The sum of three consecutive terms of an A.P. is 36 and their product is 1140.

Find the terms. (Consider the terms to be in ascanding order.)​

Answer 1

$\green{\large\underline{\sf{Solution-}}}$

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

According to statement,

Sum of three numbers = 36

$\rm :\longmapsto\:a - d + a + a + d = 36$

$\rm :\longmapsto\:3a = 36$

$\bf\implies \:a = 12$

Now,

Also it is given that

Product of three numbers = 1140

$\rm :\longmapsto\:(a - d)a(a + d) = 1140$

$\rm :\longmapsto\:(12 - d) \times 12 \times (12 + d) = 1140 \: \: \: \ \green{ \{ \because \: a = 12 \}}$

$\rm :\longmapsto\:144 - {d}^{2} = 95$

$\: \: \: \: \: \: \: \: \: \begin{gathered}\:{\underline{\boxed{\bf{\blue{{\tt \: \because \: (x + y)(x - y) = {x}^{2} - {y}^{2}}}}}}} \\ \end{gathered}$

$\rm :\longmapsto\: {d}^{2} = 144 - 95$

$\rm :\longmapsto\: {d}^{2} = 49$

$\bf\implies \:d \: = \: \pm \: 7$

Hence,

• Two cases arises

Case :- 1

When a = 12 and d = 7,

Then

$\begin{gathered}\begin{gathered}\bf\: 3 \: numbers \:in \:AP \: are -\begin{cases} &\sf{12 - 7 = 5} \\ &\sf{12}\\ &\sf{12 + 7 = 19} \end{cases}\end{gathered}\end{gathered}$

Case :- 2

When a = 12 and d = - 7

Then,

$\begin{gathered}\begin{gathered}\bf\: 3 \: numbers \:in \:AP \: are -\begin{cases} &\sf{12 + 7 = 19} \\ &\sf{12}\\ &\sf{12 - 7 = 5} \end{cases}\end{gathered}\end{gathered}$

Additional Information :-

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic sequence is,

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

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.