Q23. Prove by mathematical induction that, 41ⁿ – 14ⁿ is a multiple of 27.
Answers
Given :
In converging mirror (concave mirror),
Height of object : 5 cm.
Object distance : 15 cm.
Focal length : 30 cm.
To find :
The image distance and the size of the image formed.
Solution :
Using mirror formula that is,
» A formula which gives the relationship between image distance, object distance and focal length of a sperical mirror is known as the mirror formula .i.e.,
\boxed{\bf \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}}
v
1
+
u
1
=
f
1
where,
v denotes Image distance
u denotes object distance
f denotes focal length
By substituting all the given values in the formula,
\dashrightarrow\sf \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}⇢
v
1
+
u
1
=
f
1
\dashrightarrow\sf \dfrac{1}{v} + \dfrac{1}{ - 15} = \dfrac{1}{ - 30}⇢
v
1
+
−15
1
=
−30
1
\dashrightarrow\sf \dfrac{1}{v} - \dfrac{1}{15} = \dfrac{1}{ - 30}⇢
v
1
−
15
1
=
−30
1
\dashrightarrow\sf \dfrac{1}{v} = \dfrac{1}{ - 30} + \dfrac{1}{15}⇢
v
1
=
−30
1
+
15
1
\dashrightarrow\sf \dfrac{1}{v} = \dfrac{ - 1 + 2}{30}⇢
v
1
=
30
−1+2
\dashrightarrow\sf \dfrac{1}{v} = \dfrac{1}{30}⇢
v
1
=
30
1
\dashrightarrow\sf v = 30 \: cm⇢v=30cm
Thus, the position of the image is 30 cm.
We know that,
» The linear magnification produced by a mirror is equal to the ratio of the image distance to the object distance with a minus sign and it is equal to the ratio of image height and object height. that is,
\boxed{\bf m = \dfrac{h'}{h} = - \dfrac{v}{u}}
m=
h
h
′
=−
u
v
where,
h' denotes height of image
h denotes object height
v denotes image distance
u denotes object distance
By substituting all the given values in the formula,
\dashrightarrow\sf \dfrac{h'}{h} = - \dfrac{v}{u}⇢
h
h
′
=−
u
v
\dashrightarrow\sf \dfrac{h'}{5} = - \dfrac{ - 30}{ - 15}⇢
5
h
′
=−
−15
−30
\dashrightarrow\sf \dfrac{h'}{5} = - 2⇢
5
h
′
=−2
\dashrightarrow\sf h'= - 2 \times 5⇢h
′
=−2×5
\dashrightarrow\sf h'= -10 \: cm⇢h
′
=−10cm
Thus, the height of the image is 10 cm.
Step-by-step explanation:
41
Answer:
Let the given statement be P(n), i.e.,
P(n):41n – 14nis a multiple of 27.
It can be observed that P(n) is true for n = 1 since 411-141 =27, which is a multiple of 27.
Let P(k) be true for some positive integer k, i.e.,
41k – 14kis a multiple of 27
∴41k – 14k = 27m, where m ∈ N … (1)
We shall now prove that P(k + 1) is true whenever P(k) is true.
Consider
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.