Question

$\huge{ \sf{ \green{ \underline{ \overline{ \underline{ \overline{ \pink{ \: Question \: : - }}}}}}}}$
The sum of three numbers in G.P . is 14. If the first two terms are each increased by 1 and the third term is decreased by 1 , the resulting numbers are in A.P . Find the product of these three numbers.
a) 125
b) 64
c) 216
d) 124
$\huge{ \sf{ \green{ \underline{ \overline{ \underline{ \overline{ \pink{ \: With \: Solution \:Plz : - }}}}}}}}$​

Answer 1

Question :

The sum of three numbers in G.P. is 14 . If the first two terms are each increased by 1 and the third term is decreased by 1 , the resulting numbers are in A.P. . Find the product of these three numbers .

a. 125

b. 64

c. 216

d. 124

Solution :

Let three numbers in GP are a , ar , ar² .

The sum of three numbers in G.P . is 14

$\longrightarrow \sf a+ar+ar^2 = 14$

$\longrightarrow \sf a(1+r+r^2) = 14$

$\longrightarrow \sf a(1+r^2) = 14-ar\ \; \dashrightarrow\ (1)$

If the first two terms are each increased by 1 and the third term is decreased by 1 , the resulting numbers are in AP

So ,

➠ First term = a + 1

➠ Second term = ar + 1

➠ Third term = ar² - 1

These three terms are in AP .

$\bf \orange{If\ three\ terms\ a,b,c\ are\ in\ AP\ :}$

$\bf \green{\implies a+c=2b}$

$\longrightarrow \sf a+1+ar^2-1=2(ar+1)$

$\longrightarrow \sf a(1+r^2)=2ar+2$

$\longrightarrow \sf 14-ar=2ar+2\ \; [\ From\ (1)\ ]$

$\longrightarrow \sf 3ar=12$

$\longrightarrow \sf ar=4$

$\longrightarrow \sf a=\dfrac{4}{r}\ \; \dashrightarrow\ (2)$

Sub. eq. (2) in eq. (1) ,

$\longrightarrow \sf \dfrac{4}{r}(1+r^2) = 14-\dfrac{4}{r}\ .r$

$\longrightarrow \sf 4+4r^2=14r-4r$

$\longrightarrow \sf 4r^2-10r+4=0$

$\longrightarrow \sf 2r^2-5r+2=0$

$\longrightarrow \sf 2r^2-r-4r+2=0$

$\longrightarrow \sf r(2r-1)-2(2r-1)=0$

$\longrightarrow \sf (r-2)(2r-1)=0$

$\longrightarrow \sf r=2\ ,\ \; r=\dfrac{1}{2}$

When , r = 2 ,

➠ a = 2

When , r =  $\sf \dfrac{1}{2}$  ,

➠ a = 8

Product of three numbers :

$\longrightarrow \sf a \times ar \times ar^2$

$\longrightarrow \sf a^3 \times r^3$

$\longrightarrow \sf (ar)^3$

So ,

When r = 2 , a = 2

➠ ( 2 × 2 )³

➠ 4³

➠ 64

When r = ¹/₂ , a = 8

➠ ( 8 × ¹/₂ )³

➠ 4³

➠ 64

Product of first three numbers is 64 .

Option b   :)

Answer 2

$\Large{\underbrace{\underline{\bf{TO\: FIND\::}}}}$

• The product of three numbers which are in G.P.

$\Large{\underbrace{\underline{\bf{GIVEN\::}}}}$

• The sum of three numbers in G.P is 14.

$\Large{\underbrace{\underline{\bf{SOLUTION\::}}}}$

Let,

☛ The numbers be a, ar & ar².

$:\implies\:\bf{Sum\:=\:14}$

$:\implies\:\sf{a\:+\:ar\:+\:ar^2\:=\:14}$

$:\implies\:\sf{a\:+\:ar^2\:=\:14\:-\:ar}--(a)$

Or

$:\implies\:\sf{a\:(1\:+\:r\:+\:r^2)\:=\:14}--(b)$

$:\implies\:\sf{a\:=\:\dfrac{14}{(1\:+\:r\:+\:r^2)}}--(c)$

According to the question,

✰ (a + 1), (ar + 1) & (ar² - 1) are in A.P.

As we know that,

➳ 2 (ar + 1) = (a + 1) + (ar² - 1)

➳ 2ar + 2 = a + 1 + ar² - 1

➳ 2ar + 2 = a + ar²

✳ Putting the value of equation (a), in the above equation given as,

➳ 2ar + 2 = 14 - ar

➳ 2ar + ar = 14 - 2

➳ 3ar = 12

➳ ar = $\sf{\dfrac{12}{3}}$

➳ ar = 4

➳ a = $\sf{\dfrac{4}{r}}$

➳ $\sf{\dfrac{14}{(1\:+\:r\:+\:r^2)}}$ = $\sf{\dfrac{4}{r}}$

➳ 14 = $\sf{\dfrac{4}{r}}$ × (1 + r + r²)

➳ 14r = 4 + 4r + 4r²

➳ 4r² + 4r - 14r + 4 = 0

➳ 4r² - 10r + 4 = 0

➳ 2r² - 5r + 2 = 0

➳ 2r² - 4r - r + 2 = 0

➳ 2r (r - 2) -1 (r - 2) = 0

➳ (2r - 1) (r - 2) = 0

➳ 2r - 1 = 0 or r - 2 = 0

➳ 2r = 1 or r = 2

➳ r = $\bf{\dfrac{1}{2}}$ or r = 2

✳ Put 'r = $\bf{\dfrac{1}{2}}$' in the equation (b) given as,

➵ a $\bigg[$1 + $\rm{\dfrac{1}{2}}$ + $\rm{\Big(\dfrac{1}{2}\Big)^2}\:\bigg]$ = 14

➵ a $\bigg[$1 + $\rm{\dfrac{1}{2}}$ + $\rm{\dfrac{1}{4}}\:\bigg]$ = 14

➵ a $\rm{\bigg[\dfrac{4\:+\:2\:+\:1}{4}}\:\bigg]$ = 14

➵ a $\rm{\bigg[\dfrac{7}{4}}\:\bigg]$ = 14

➵ a = 14 × $\sf{\dfrac{4}{7}}$

➵ a = 2 × 4

➵ a = 8

And

➵ ar = 8 × $\sf{\dfrac{1}{2}}$ = 4

And

➵ ar² = 8 × $\sf{\Big(\dfrac{1}{2}\Big)^2}$ = 8 × $\sf{\dfrac{1}{4}}$ = 2

Therefore,

• 2, 4 & 8 are in G.P.

∴ The product of these three numbers are given as,

$\longrightarrow\:\sf{2\times{4}\times{8}}$

$\longrightarrow\:\bf\pink{64}$

[Note ➝ If we put 'r = 2', then the result is same as above.]