प्रtham n-विषम संख्याओं का योगफला hain
Answers
Answer:
The possible subsets are ϕ,{a},{b},{c},{a,b},{b,c},{a,c},{a,b,c}.The possible subsets are ϕ,{a},{b},{c},{a,b},{b,c},{a,c},{a,b,c}.sin3A=cos(A−10
o
)⇒cos(90
o
−3A)=cos(A−10
o
)
⇒90
o
+10
o
=3A+A⇒100
o
=4A⇒A=25
o
Therefore, Answer is 25
osin3A=cos(A−10
o
)⇒cos(90
o
−3A)=cos(A−10
o
)
⇒90
o
+10
o
=3A+A⇒100
o
=4A⇒A=25
o
Therefore, Answer is 25
oGiven, sinA=
5
3
=
h
p
∴b=
25−9
=
16
=4
secA=
b
h
=
4
5
,
tanA=
b
p
=
4
3
Therefore, sec
2
A−tan
2
A=(
4
5
)
2
−(
4
3
)
2
=
16
25
−
16
9
=
16
16
=1The possible subsets are ϕ,{a},{b},{c},{a,b},{b,c},{a,c},{a,b,c}.The possible subsets are ϕ,{a},{b},{c},{a,b},{b,c},{a,c},{a,b,c}.The given equations are ab+bc=44 ..........(i)
And, ac+bc=23 .........(ii)
⇒c(a+b)=23, which is a prime number
So, the two factors must be 1 & 23.
c=1 and $$a + b = 23$$
⇒b=23−a
Put these values in (i), b(a+c)=44
⇒(23−a)(a+1)=44
⇒23a−a
2
+23−a=44
⇒a
2
−22a+21=0
⇒a
2
−21a−a+21=0
⇒a(a−21)−1(a−21)=0
⇒(a−21)(a−1)=0
⇒a=1,21
For, a=1,b=22
For a=21,b=2
∴b=22,2
The solution sets are (1,22,1);(21,2,1).
Hence, the number of solutions sets =2.Given, sinA=
5
3
=
h
p
∴b=
25−9
=
16
=4
secA=
b
h
=
4
5
,
tanA=
b
p
=
4
3
Therefore, sec
2
A−tan
2
A=(
4
5
)
2
−(
4
3
)
2
=
16
25
−
16
9
=
16
16
=1Let a be the first term, r be the common ratio and l be the last term of a GP containing n terms.
Then,
(pth term from the beginning)×(pth term from the end)
=(pth term from the beginning)×{(n−p+1)the term from teh beginning}
=T
p
×T
(n−p+1)
=ar
p−1
×ar
(n−p+1)−1
=ar
(p−1)
×ar
(n−p)
=a×(ar
n−1
)=(T
1
×T
n
)=(first term × last term).Given, sinA=
5
3
=
h
p
∴b=
25−9
=
16
=4
secA=
b
h
=
4
5
,
tanA=
b
p
=
4
3
Therefore, sec
2
A−tan
2
A=(
4
5
)
2
−(
4
3
)
2
=
16
25
−
16
9
=
16
16
=1sin3A=cos(A−10
o
)⇒cos(90
o
−3A)=cos(A−10
o
)
⇒90
o
+10
o
=3A+A⇒100
o
=4A⇒A=25
o
Therefore, Answer is 25
osin3A=cos(A−10
o
)⇒cos(90
o
−3A)=cos(A−10
o
)
⇒90
o
+10
o
=3A+A⇒100
o
=4A⇒A=25
o
Therefore, Answer is 25
oThe given equations are ab+bc=44 ..........(i)
And, ac+bc=23 .........(ii)
⇒c(a+b)=23, which is a prime number
So, the two factors must be 1 & 23.
c=1 and $$a + b = 23$$
⇒b=23−a
Put these values in (i), b(a+c)=44
⇒(23−a)(a+1)=44
⇒23a−a
2
+23−a=44
⇒a
2
−22a+21=0
⇒a
2
−21a−a+21=0
⇒a(a−21)−1(a−21)=0
⇒(a−21)(a−1)=0
⇒a=1,21
For, a=1,b=22
For a=21,b=2
∴b=22,2
The solution sets are (1,22,1);(21,2,1).
Hence, the number of solutions sets =2.The possible subsets are ϕ,{a},{b},{c},{a,b},{b,c},{a,c},{a,b,c}.Let a be the first term, r be the common ratio and l be the last term of a GP containing n terms.
Then,
(pth term from the beginning)×(pth term from the end)
=(pth term from the beginning)×{(n−p+1)the term from teh beginning}
=T
p
×T
(n−p+1)
=ar
p−1
×ar
(n−p+1)−1
=ar
(p−1)
×ar
(n−p)
=a×(ar
n−1
)=(T
1
×T The possible subsets are ϕ,{a},{b},{c},{a,b},{b,c},{a,c},{a,b,c}.The possible subsets are ϕ,{a},{b},{c},{a,b},{b,c},{a,c},{a,b,c}.The possible subsets are ϕ,{a},{b},{c},{a,b},{b,c},{a,c},{a,b,c}.The possible subsets are ϕ,{a},{b},{c},{a,b},{b,c},{a,c},{a,b,c}.The possible subsets are ϕ,{a},{b},{c},{a,b},{b,c},{a,c},{a,b,c}.The possible subsets are ϕ,{a},{b},{c},{a,b},{b,c},{a,c},{a,b,c}.The possible subsets are ϕ,{a},{b},{c},{a,b},{b,c},{a,c},{a,b,c}.The possible subsets are ϕ,{a},{b},{c},{a,b},{b,c},{a,c},{a,b,c}.The possible subsets are ϕ,{a},{b},{c},{a,b},{b,c},{a,c},{a,b,c}.The possible subsets are ϕ,{a},{b},{c},{a,b},{b,c},{a,c},{a,b,c}.The possible subsets are ϕ,{a},{b},{c},{a,b},{b,c},{a,c},{a,b,c}.The possible subsets are ϕ,{a},{b},{c},{a,b},{b,c},{a,c},{a,b,c}.
n
)=(first term × last term).The possible subsets are ϕ,{a},{b},{c},{a,b},{b,c},{a,c},{a,b,c}.