To find the sum of a number of terms in Arithmetical Progression:
Let a=first term, d = common difference, l=t n =last term, s = required sum. Then, Writing the series in the reverse order, Adding together the two series, ..
Let a=first term, d = common difference, l=t n =last term, s = required sum. Then, Writing the series in the reverse order, Adding together the two series, ..To find the sum of n terms of a GP
Let a = First term, r = common ratio, n = number of terms. Multiply both sides of (i) by r, the common ratio. Subtracting (ii) from (i), we get ..
Let a = First term, r = common ratio, n = number of terms. Multiply both sides of (i) by r, the common ratio. Subtracting (ii) from (i), we get ..Harmonic Progression (H.P.)
A sequence of numbers is said to form a harmonic progression if their reciprocals form an arithmetic progressio..
A sequence of numbers is said to form a harmonic progression if their reciprocals form an arithmetic progressio..Suggested answer:
n t h term of (1,2,3,....) is n Subtracting (ii) from (i), we get ..
n t h term of (1,2,3,....) is n Subtracting (ii) from (i), we get ..Factorial Questions
Question 1 - Question: Find n. Answer: ..
Question 1 - Question: Find n. Answer: ..Arithmetic Geometric
Arithmetic Geometric Series - A series of the form a + (a + d)r + (a + 2d)r 2 + ... is called an Arithmetic-Geometric series. In the series if we put we get GP and if we put r = 1, we get an AP. To find the sum to the series Subtracting (ii) from (i), we g..
Arithmetic Geometric Series - A series of the form a + (a + d)r + (a + 2d)r 2 + ... is called an Arithmetic-Geometric series. In the series if we put we get GP and if we put r = 1, we get an AP. To find the sum to the series Subtracting (ii) from (i), we g..Question 4
Question: When n = 5 and r = 2, find the values of Answer: i) ii) ..
Question: When n = 5 and r = 2, find the values of Answer: i) ii) ..Question 3
Question: Answer: = 5 + 10 + 10 + 5 + 1 = 31 = RH..
Question: Answer: = 5 + 10 + 10 + 5 + 1 = 31 = RH..Area of a Triangle
We have already learnt in the previous class that the area of triangle whose vertices are (x 1 , y 1 ), (x 2 , y 2 ), (x 3 , y 3 ) is given by Hence area of a triangle having vertices at (x 1 , y 1 ), (x 2 , y 2 ) and (x 3 , y 3 ) is given by..
We have already learnt in the previous class that the area of triangle whose vertices are (x 1 , y 1 ), (x 2 , y 2 ), (x 3 , y 3 ) is given by Hence area of a triangle having vertices at (x 1 , y 1 ), (x 2 , y 2 ) and (x 3 , y 3 ) is given by..Suggested answer:
The given equations are 2x - y + z = -3 3x - 0.y - z = - 8 2x + 6y + 0.z= 2 = 2(6) +1(2) + 1(18) = 12 +2 + 18 = 32 The system has a unique solutions. A 1 1 = (0 + 6) = 6, A 1 2 = -(0 + 2) = -2, A 1 3 = 18 A 2 1 =..
The given equations are 2x - y + z = -3 3x - 0.y - z = - 8 2x + 6y + 0.z= 2 = 2(6) +1(2) + 1(18) = 12 +2 + 18 = 32 The system has a unique solutions. A 1 1 = (0 + 6) = 6, A 1 2 = -(0 + 2) = -2, A 1 3 = 18 A 2 1 =.. Result
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