let us assume a be the first term and d the common difference.
According to Question,
mth term of an Arithmetic Progression (AP) is 1/n
Use the Arithmetic Progression formula for mth term,
? a + ( m - 1 ) x d = tm
? a + ( m - 1 ) x d = 1/n
? a + md - d = 1/n .......................(1)
Use the Arithmetic Progression formula for nth term,
? a + ( n - 1 ) x d = tn
? a + ( n - 1 ) x d = 1/m
? a + nd - d = 1/m.........................(2)
Subtracts Equation (1) from Equation (2) , we will get
? a + nd - d - (a + md - d) = 1/m - 1/n
? a + nd - d - (a + md - d) = 1/m - 1/n
? a + nd - d - a - md + d = 1/m - 1/n
? nd - md = 1/m - 1/n
? d(n - m) = 1/m - 1/n
? d(n - m) = (n - m)/mn
? d = 1/mn.......................................(3)
Put the value of d in Equation (1), we will get,
? a + md - d = 1/n
? a + ( m x 1/mn ) - 1/mn = 1/n
? a + 1/n - 1/mn = 1/n
? a = 1/n - 1/n + 1/mn
? a = 1/mn.....................................(4)
Now according to question,
Sum of mn terms = mn/2(2a + ( mn - 1) x d)
Smn = mn/2(2a + ( mn - 1) x d)
Put the value of a and d in above equation, we will get,
Smn = mn/2 [ 2 x 1/mn + ( mn - 1) x 1/mn ]
Smn = mn/2 [2/mn + 1 - 1/mn ]
Smn = mn/2 [ 1 + 1/mn ]
Smn = mn/2 [ ( mn + 1 )/mn ]
Smn = mn/2 (1 + mn)/mn
Smn = 1/2 (1 + mn)
Smn = (1 + mn)/2 = (mn + 1)/2
Let the two consecutive numbers are N and N+2
According to the question
N (N + 2) = 19043
? N2 + 2N - 19043 = 0
? N2 + 139N - 137N - 19043= 0
? N( N + 139) - 137 ( N - 137 ) = 0
? N( N + 139) ( N - 137) = 0
? N = 137 and N = - 139
? N = 137
A will reach at starting point in 5 * 2 / 5 = 2 hours ;
B will reach at starting point in 5 / 3 hours ;
C will reach at starting point in 5 / 2 hours ;
Then, on the starting point all three will meet after the L.C.M. of 2, 5 / 3, 5 / 2, 10 / 1 = 10 hours.
Circumference = 2?r
= 2 x (22 / 7) x 70 cm
= 440 cm
Distance travelled in 10 revolutions = 440 x 10 cm
= 4400 cm
= 44 m
? Speed = distance / time
= 44 / 5 m/sec
= (44 / 5) x (18 / 5) km/hr
= 31.68 km/hr
Regarding all copies of the same book as one book, we have only 5 books. These 5 books can be arranged in 5! ways. But all copies of the same book being identical can be arranged in only one way.
? Required number = 5! x 1! x 1! x 1! x 1! = 120
We known that,
LCM of fractions = (LCM of numerators) / (HCF of denominators)
? Required LCM = (LCM of 1, 2, 5, and 4) / (HCF of 3, 9, 6 and 27)
? = 20/3
The speed of A and B are in the ratio 11 : 8.
Let, speeds be 11s and 8s (in m/sec).
Let, race be of P meter.
Then, time taken by A to run P meter is same as that of B to run (P - 120) meter.
? P / 11 = (P - 120) / 8
? P / 11 = (P - 120 / 8
? 8P = 11 x (P - 120 )
? 8P = 11P - 120 x 11
? 11P - 8P = 120 x 11
? 3P = 11 × 120
? P = 440.
? = 124.35% of 8096
= (8096 x 124.35)/100 = 1006737.6/100
= 10067.376 = 10000
log105 = log10(10/2)
= log1010 - log102
= 1-0.3010
= 0.6990
Friday will fall on 3, 10, 17, 24, 31
So, it will be 5th Friday on 31 st
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