Question

Find four consecutive terms in an A.P. whose sum is 72and the ratio of product of 1st and 4th terms to product 2nd and 3rd terms 9:10.​

Answer 1

$\bold{\Huge{\underline{\boxed{\bf{\red{ANSWER\::}}}}}}$

$\bold{\Large{\underline{\sf{\green{Given\::}}}}}$

In an A.P. whose sum is 72 and the ratio of product of 1st term and 4th term to product 2nd term and 3rd terms 9:10.

$\bold{\Large{\underline{\sf{\pink{To\:find\::}}}}}$

The four consecutive terms in an A.P.

$\bold{\Large{\underline{\sf{\orange{Explanation\::}}}}}$

Let the four terms of an A.P. is ;

$\bold{ \begin{cases}\bf{First\:term=a-3d}\\ \bf{Second\:term=a-d}\\ \bf{Third\:term=a+d}\\ \bf{Fourth\:term=a+3d}\end{cases}}$

According to the question:

→ a-3d + a-d + a+d + a+3d = 72

→ $\bf{a\cancel{-3d}+a\cancel{-d}+a+\cancel{d}+a+\cancel{3d}=72}$

→ 4a = 72

→ a = $\bold{\cancel{\frac{72}{4} }}$

→ a = 18

Now,

The ratio of product of first and fourth terms to product second and third terms 9:10.

→ $\bf{\frac{(a-3d)(a+3d)}{(a-d)(a+d)} =\frac{9}{10} }$

→ $\bold{\frac{a^{2} +3ad-3ad-9d^{2} }{a^{2}+ad-ad-d^{2}  } =\frac{9}{10} }$

→ $\bold{\frac{a^{2} -9d^{2} } {a^{2} - d^{2}  } =\frac{9}{10} }$

→ 10a² - 90d² = 9a² - 9d²

→ 10a² - 9a² = -9d² + 90d²

→ a² = 81d²

→ a = √81d²

→ a = ±9d

→ d = ±$\bf\frac{a}{9} }$

→ d = ±$\bf\frac{18}{9} }$     [Putting the value of a]

→ d = ±$\bf{\cancel{\frac{18}{9} }}$

→ d = ±2

Now,

Putting the value of d = +2 & a = 18, we get;

• The four consecutive terms in an A.P.

→ First term = 18 - 3(2) = 18 - 6 = 12

→ Second term = 18 - 2 = 16

→ Third term = 18 +2 = 20

→ Fourth term = 18 + 3(2) = 18 + 6 = 24

&

Putting the value of d = -2 and a = 18, we get;

→ First term= 18 - 3(-2) = 18 - (-6) = 18 + 6 = 24

→ Second term= 18 - (-2) = 18 + 2 = 20

→ Third term= 18 + (-2) = 18 - 2 = 16

→ Fourth term= 18 + 3(-2) = 18 - 6 = 12

Answer 2

$\huge\underline\blue{AnSweR}$...

In an A.P. whose sum is 72 and the ratio of product of 1st term and 4th term to product 2nd term and 3rd terms 9:10.

\bold{\Large{\underline{\sf{\pink{To\:find\::}}}}}

Tofind:

The four consecutive terms in an A.P.

\bold{\Large{\underline{\sf{\orange{Explanation\::}}}}}

Explanation:

Let the four terms of an A.P. is ;

\begin{lgathered}\bold{ \begin{cases}\bf{First\:term=a-3d}\\ \bf{Second\:term=a-d}\\ \bf{Third\:term=a+d}\\ \bf{Fourth\:term=a+3d}\end{cases}}\end{lgathered}

⎩

⎪

⎪

⎪

⎨

⎪

⎪

⎪

⎧

Firstterm=a−3d

Secondterm=a−d

Thirdterm=a+d

Fourthterm=a+3d

According to the question:

→ a-3d + a-d + a+d + a+3d = 72

→ \bf{a\cancel{-3d}+a\cancel{-d}+a+\cancel{d}+a+\cancel{3d}=72}a

−3d

+a

−d

+a+

d

+a+

3d

=72

→ 4a = 72

→ a = \bold{\cancel{\frac{72}{4} }}

4

72

→ a = 18

Now,

The ratio of product of first and fourth terms to product second and third terms 9:10.

→ \bf{\frac{(a-3d)(a+3d)}{(a-d)(a+d)} =\frac{9}{10} }

(a−d)(a+d)

(a−3d)(a+3d)

=

10

9

→

→

→ 10a² - 90d² = 9a² - 9d²

→ 10a² - 9a² = -9d² + 90d²

→ a² = 81d²

→ a = √81d²

→ a = ±9d

→ d = ±

→ d = ± [Putting the value of a]

→ d = ±\bf{\cancel{\frac{18}{9} }}

9

18

→ d = ±2

Now,

Putting the value of d = +2 & a = 18, we get;

The four consecutive terms in an A.P.

→ First term = 18 - 3(2) = 18 - 6 = 12

→ Second term = 18 - 2 = 16

→ Third term = 18 +2 = 20

→ Fourth term= 18 + 3(-2) = 18 - 6 = 12