Question

Date
find four consecutive terms in an A.p. such that
the sum of the middle two terms is 18 and
product of the two end tems is 45.
(Assume the four consecutive
term in hipeare
a-d, a, a+d ,a+2d?​

Answer 1

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

Let the four consecutive numbers be a - d, a, a + d, a + 2d.

According to statement,

Sum of middle two terms = 18

$\rm :\longmapsto\:a + a + d = 18$

$\rm :\longmapsto\:2a + d = 18$

$\rm :\longmapsto\:d = 18 - 2a - - - (1)$

According to statement,

Product of two end terms = 45

$\rm :\longmapsto\:(a - d)(a + 2d) = 45$

On substituting the value of d, we get

$\rm :\longmapsto\:(a - 18 + 2a)(a + 36 - 4a) = 45$

$\rm :\longmapsto\:(3a - 18)(36 - 3a) = 45$

$\rm :\longmapsto\:9(a - 6)(12 - a) = 45$

$\rm :\longmapsto\:(a - 6)(12 - a) = 5$

$\rm :\longmapsto\:12a - {a}^{2} - 72 + 6a = 5$

$\rm :\longmapsto\:18a - {a}^{2} - 72= 5$

$\rm :\longmapsto\:{a}^{2} - 18a + 77 = 0$

$\rm :\longmapsto\:{a}^{2} - 11a - 7a + 77 = 0$

$\rm :\longmapsto\:a(a - 11) - 7(a - 11) = 0$

$\rm :\longmapsto\:(a - 11)(a - 7) = 0$

$\bf\implies \:a = 11 \: \: \: or \: \: \: a = 7$

Now,

When a = 11, d = 18 - 2a = 18 - 22 = - 4

So, numbers are 15, 11, 7, 3

When a = 7, d = 18 - 2a = 18 - 14 = 4

So, numbers are 3, 7, 11, 15

Additional Information :-

↝ 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.

and

↝ Sum of n  terms of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• Sₙ is the sum of n terms of AP.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.