Question

Sum of four terms in an GP is 312 . The sum oh fiurst and fourth telrm is 252 . Finsd the productr of second and thierd term ​

Answer 1

Correct Question :

The Sum of four consecutive terms in a GP is 312 . The sum of first and fourth term is 252 . Find the product of second and third term

Given :

• Sum of four terms in GP is 312

• Sum of first and third term is 252

To Find :

• The product of Second and third term

Solution :

The general consecutive terms of GP are a , ar , ar² , ar³ , ar⁴ ,.......arⁿ

Where ,

• a is first term
• r is common ratio

Now let the four terms be ,

• a , ar , ar² , ar³

Sum of first and third term is 252

$: \implies \rm \: a + ar {}^{3} = 252 \\ \\ : \implies \rm \: a(1 + {r}^{3} ) = 252 \longrightarrow \: eq(1)\\ \\ : \implies \rm \:a [(1) {}^{3} + ( {r}^{3} ) ] = 252$

It is in the form of a³ + b³ ,

${\boxed {\rm{ {a}^{3} + {b}^{3} = (a + b)( {a}^{2} - ab + {b}^{2} )}}}$

Applying the formula ,

$: \implies \rm \: a[(1 + r)(1 - r + {r}^{2} ) ] = 252 \longrightarrow \: eq(2)$

We have Sum of the four consecutive terms as 312 ,

$: \implies \rm a + ar + ar {}^{2} + ar {}^{3} = 312 \\ \\ : \implies \rm \: a(1 + r + {r}^{2} + {r}^{3} ) = 312 \\ \\ : \implies \rm \: a[ (1 + r)(1 + r {}^{2}) ] = 312 \longrightarrow \: eq(3)$

Dividing eq(2) by eq(3) ,

$: \implies \rm \: \dfrac{a[(1 + r)(1 - r + {r}^{2}) ]}{a[ (1 + r)(1 + {r}^{2}) ]} = \dfrac{312}{252} \\ \\ : \implies \rm \: \frac{ \cancel{a[ (1 + r)} (1 + {r}^{2}) ]}{ \cancel{a[ (1 + r)} (1 - r + {r}^{2}) ]} = \frac{26}{21} \\ \\ : \implies \rm \: \frac{(1 + {r}^{2}) }{(1 - r+ {r}^{2} )} = \frac{26}{21} \\ \\ : \implies \rm \: 21[ (1 + {r}^{2}) ] =26 [ (1 - r + {r}^{2}) ] \\ \\ : \implies \rm \: 21 + 21 {r}^{2} = 26 - 26r + 26 {r}^{2}$

$: \implies \rm \: 21 + 21 {r}^{2} - 26 + 26r - 26 {r}^{2} = 0 \\ \\ : \implies \rm \: - 5 - 5 {r}^{2} + 26r = 0 \\ \\ :\implies \rm \: 5r {}^{2} - 26r + 5 = 0 \\ \\ : \implies \rm \: 5 {r}^{2} - 25r - r + 5 = 0 \\ \\ : \implies \rm \: 5r(r - 5) - 1(r - 5) = 0 \\ \\ : \implies \rm \: (5r - 1)(r - 5) = 0$

$:\implies \rm \: r = \dfrac{1}{5} \: (or) \: r = 5$

When

(1) r = 1/5

From eq(1) ,

$: \implies \rm \: a(1 + r {}^{3} ) = 252 \\ \\ :\implies \rm \: a \bigg[ 1 + \bigg( \frac{1}{5} \bigg)^{3} \bigg] = 252 \\ \\ : \implies \rm \: a \bigg[ 1 + \frac{1}{125} \bigg] = 252 \\ \\ :\implies \rm \: a \bigg[ \frac{126}{125} \bigg] = 252 \\ \\ :\implies \rm \: a = 252 \times \frac{125}{126} \\ \\ : \implies \rm \: a = 250 \: \pink\bigstar$

When

(2) r = 5

From eq(1) ,

$: \implies \rm \: a[1 + (5) {}^{3} ] = 252 \\ \\ : \implies \rm \: a(1 + 125) = 252 \\ \\ :\implies \rm \: a(126) = 252 \\ \\ :\implies \rm \: a = \frac{252}{126} \\ \\ :\implies \rm \: a = 2 \: \: \pink\bigstar$

When

(1) a = 2 , r = 5 The terms are ,

$\rm \: a = 2 \\ \\ \rm ar = 2(5) = 10 \\ \\ \rm \: a {r}^{2} = 2( {5}^{2} ) = 50 \\ \\ \rm \: a {r}^{3} = 2( {5}^{3} ) = 250$

Then , Then the Product of second and third term = 50 × 10 = 500

(2) When a = 250 , r = 1/5 The terms are ,

$\rm \: a = 250 \\ \\ \rm \: ar = 250 \bigg( \frac{1}{5} \bigg) = 50 \\ \\ \rm \: a {r}^{2} = 250 \bigg( \frac{1}{25} \bigg) = 10 \\ \\ \rm \: a {r}^{3} = 250 \bigg( \frac{1}{125} \bigg) = 2$

Then , The product of second and third terms = 50 × 10 = 500

In both cases the product of second and third term is 500.

∴ The required answer is 500

Answer 2

Answer:

Correct Question :

The Sum of four consecutive terms in a GP is 312 . The sum of first and fourth term is 252 . Find the product of second and third term

Given :

Sum of four terms in GP is 312

Sum of first and third term is 252

To Find :

The product of Second and third term

Solution :

The general consecutive terms of GP are a , ar , ar² , ar³ , ar⁴ ,.......arⁿ

Where ,

a is first term

r is common ratio

Now let the four terms be ,

a , ar , ar² , ar³

Sum of first and third term is 252

\begin{gathered} : \implies \rm \: a + ar {}^{3} = 252 \\ \\ : \implies \rm \: a(1 + {r}^{3} ) = 252 \longrightarrow \: eq(1)\\ \\ : \implies \rm \:a [(1) {}^{3} + ( {r}^{3} ) ] = 252 \end{gathered}

:⟹a+ar

3

=252

:⟹a(1+r

3

)=252⟶eq(1)

:⟹a[(1)

3

+(r

3

)]=252

It is in the form of a³ + b³ ,

{\boxed {\rm{ {a}^{3} + {b}^{3} = (a + b)( {a}^{2} - ab + {b}^{2} )}}}

a

3

+b

3

=(a+b)(a

2

−ab+b

2

)

Applying the formula ,

: \implies \rm \: a[(1 + r)(1 - r + {r}^{2} ) ] = 252 \longrightarrow \: eq(2):⟹a[(1+r)(1−r+r

2

)]=252⟶eq(2)

We have Sum of the four consecutive terms as 312 ,

\begin{gathered} : \implies \rm a + ar + ar {}^{2} + ar {}^{3} = 312 \\ \\ : \implies \rm \: a(1 + r + {r}^{2} + {r}^{3} ) = 312 \\ \\ : \implies \rm \: a[ (1 + r)(1 + r {}^{2}) ] = 312 \longrightarrow \: eq(3)\end{gathered}

:⟹a+ar+ar

2

+ar

3

=312

:⟹a(1+r+r

2

+r

3

)=312

:⟹a[(1+r)(1+r

2

)]=312⟶eq(3)

Dividing eq(2) by eq(3) ,

\begin{gathered} : \implies \rm \: \dfrac{a[(1 + r)(1 - r + {r}^{2}) ]}{a[ (1 + r)(1 + {r}^{2}) ]} = \dfrac{312}{252} \\ \\ : \implies \rm \: \frac{ \cancel{a[ (1 + r)} (1 + {r}^{2}) ]}{ \cancel{a[ (1 + r)} (1 - r + {r}^{2}) ]} = \frac{26}{21} \\ \\ : \implies \rm \: \frac{(1 + {r}^{2}) }{(1 - r+ {r}^{2} )} = \frac{26}{21} \\ \\ : \implies \rm \: 21[ (1 + {r}^{2}) ] =26 [ (1 - r + {r}^{2}) ] \\ \\ : \implies \rm \: 21 + 21 {r}^{2} = 26 - 26r + 26 {r}^{2} \end{gathered}

:⟹

a[(1+r)(1+r

2

)]

a[(1+r)(1−r+r

2

)]

=

252

312

:⟹

a[(1+r)

(1−r+r

2

)]

a[(1+r)

(1+r

2

)]

=

21

26

:⟹

(1−r+r

2

)

(1+r

2

)

=

21

26

:⟹21[(1+r

2

)]=26[(1−r+r

2

)]

:⟹21+21r

2

=26−26r+26r

2

\begin{gathered} : \implies \rm \: 21 + 21 {r}^{2} - 26 + 26r - 26 {r}^{2} = 0 \\ \\ : \implies \rm \: - 5 - 5 {r}^{2} + 26r = 0 \\ \\ :\implies \rm \: 5r {}^{2} - 26r + 5 = 0 \\ \\ : \implies \rm \: 5 {r}^{2} - 25r - r + 5 = 0 \\ \\ : \implies \rm \: 5r(r - 5) - 1(r - 5) = 0 \\ \\ : \implies \rm \: (5r - 1)(r - 5) = 0\end{gathered}

:⟹21+21r

2

−26+26r−26r

2

=0

:⟹−5−5r

2

+26r=0

:⟹5r

2

−26r+5=0

:⟹5r

2

−25r−r+5=0

:⟹5r(r−5)−1(r−5)=0

:⟹(5r−1)(r−5)=0

:\implies \rm \: r = \dfrac{1}{5} \: (or) \: r = 5:⟹r=

5

1

(or)r=5

When

(1) r = 1/5

From eq(1) ,

\begin{gathered}: \implies \rm \: a(1 + r {}^{3} ) = 252 \\ \\ :\implies \rm \: a \bigg[ 1 + \bigg( \frac{1}{5} \bigg)^{3} \bigg] = 252 \\ \\ : \implies \rm \: a \bigg[ 1 + \frac{1}{125} \bigg] = 252 \\ \\ :\implies \rm \: a \bigg[ \frac{126}{125} \bigg] = 252 \\ \\ :\implies \rm \: a = 252 \times \frac{125}{126} \\ \\ : \implies \rm \: a = 250 \: \pink\bigstar\end{gathered}

:⟹a(1+r

3

)=252

:⟹a[1+(

5

1

)

3

]=252

:⟹a[1+

125

1

]=252

:⟹a[

125

126

]=252

:⟹a=252×

126

125

:⟹a=250★

When

(2) r = 5

From eq(1) ,

\begin{gathered} : \implies \rm \: a[1 + (5) {}^{3} ] = 252 \\ \\ : \implies \rm \: a(1 + 125) = 252 \\ \\ :\implies \rm \: a(126) = 252 \\ \\ :\implies \rm \: a = \frac{252}{126} \\ \\ :\implies \rm \: a = 2 \: \: \pink\bigstar\end{gathered}

:⟹a[1+(5)

3

]=252

:⟹a(1+125)=252

:⟹a(126)=252

:⟹a=

126

252

:⟹a=2★

When

(1) a = 2 , r = 5 The terms are ,

\begin{gathered} \rm \: a = 2 \\ \\ \rm ar = 2(5) = 10 \\ \\ \rm \: a {r}^{2} = 2( {5}^{2} ) = 50 \\ \\ \rm \: a {r}^{3} = 2( {5}^{3} ) = 250\end{gathered}

a=2

ar=2(5)=10

ar

2

=2(5

2

)=50

ar

3

=2(5

3

)=250

Then , Then the Product of second and third term = 50 × 10 = 500

(2) When a = 250 , r = 1/5 The terms are ,

\begin{gathered} \rm \: a = 250 \\ \\ \rm \: ar = 250 \bigg( \frac{1}{5} \bigg) = 50 \\ \\ \rm \: a {r}^{2} = 250 \bigg( \frac{1}{25} \bigg) = 10 \\ \\ \rm \: a {r}^{3} = 250 \bigg( \frac{1}{125} \bigg) = 2\end{gathered}

a=250

ar=250(

5

1

)=50

ar

2

=250(

25

1

)=10

ar

3

=250(

125

1

)=2

Then , The product of second and third terms = 50 × 10 = 500

In both cases the product of second and third term is 500.

∴ The required answer is 500