a) An article is sold from Jaipur (Rajasthan) to Indore
(M.P.) for 5000 and then from Indore to Bhopal.
If the rate of tax under GST system is 18% and the
profit made by the dealer in Indore is 2,000, find :
i) Net GST payable by the dealer in Indore.
ii) C.P. for the dealer in Bhopal.
Answers
Explanation:
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⟶a+ar+ar
2
=14
\longrightarrow \sf a(1+r+r^2) = 14⟶a(1+r+r
2
)=14
\longrightarrow \sf a(1+r^2) = 14-ar\ \; \dashrightarrow\ (1)⟶a(1+r
2
)=14−ar ⇢ (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\ :}If three terms a,b,c are in AP :
\bf \green{\implies a+c=2b}⟹a+c=2b
\longrightarrow \sf a+1+ar^2-1=2(ar+1)⟶a+1+ar
2
−1=2(ar+1)
\longrightarrow \sf a(1+r^2)=2ar+2⟶a(1+r
2
)=2ar+2
\longrightarrow \sf 14-ar=2ar+2\ \; [\ From\ (1)\ ]⟶14−ar=2ar+2 [ From (1) ]
\longrightarrow \sf 3ar=12⟶3ar=12
\longrightarrow \sf ar=4⟶ar=4
\longrightarrow \sf a=\dfrac{4}{r}\ \; \dashrightarrow\ (2)⟶a=
r
4
⇢ (2)
Sub. eq. (2) in eq. (1) ,
\longrightarrow \sf \dfrac{4}{r}(1+r^2) = 14-\dfrac{4}{r}\ .r⟶
r
4
(1+r
2
)=14−
r
4
.r
\longrightarrow \sf 4+4r^2=14r-4r⟶4+4r
2
=14r−4r
\longrightarrow \sf 4r^2-10r+4=0⟶4r
2
−10r+4=0
\longrightarrow \sf 2r^2-5r+2=0⟶2r
2
−5r+2=0
\longrightarrow \sf 2r^2-r-4r+2=0⟶2r
2
−r−4r+2=0
\longrightarrow \sf r(2r-1)-2(2r-1)=0⟶r(2r−1)−2(2r−1)=0
\longrightarrow \sf (r-2)(2r-1)=0⟶(r−2)(2r−1)=0
\longrightarrow \sf r=2\ ,\ \; r=\dfrac{1}{2}⟶r=2 , r=
2
1
When , r = 2 ,
➠ a = 2
When , r = \sf \dfrac{1}{2}
2
1
,
➠ a = 8
Product of three numbers :
\longrightarrow \sf a \times ar \times ar^2⟶a×ar×ar
2
\longrightarrow \sf a^3 \times r^3⟶a
3
×r
3
\longrightarrow \sf (ar)^3⟶(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 :)
thank you