In the following APs, find the missing terms in the boxes : [], 38, [], [], [], -22 Here, a₂ = 38 and a₆ = - 22. To find: a₁, a₃, a₄ and a₅ Given that: a₂ = a + (2 -1)d = 38 ⇒ a + d = 38 ⇒ a = 38 - d and a₆ = - 22 ⇒ a + 5d = - 22 Putting the value of a from equation (1), we get 38 - d + 5d = - 22 ⇒Read more
In the following APs, find the missing terms in the boxes : [], 38, [], [], [], -22
Here, a₂ = 38 and a₆ = – 22. To find: a₁, a₃, a₄ and a₅
Given that: a₂ = a + (2 -1)d = 38
⇒ a + d = 38
⇒ a = 38 – d
and a₆ = – 22 ⇒ a + 5d = – 22
Putting the value of a from equation (1), we get
38 – d + 5d = – 22 ⇒ d = – 15
Putting the value of a from equation (1), we get
a = 38 – (-15) = 53
Therefore, a₁ = a = 53
a₃ = a + 2d = 53 +2(-15) = 23
a₄ = a + 3d = 53 + 3 (-15) = 8
a₅ = a + 4d = 53 + 4 (-15) = – 7
Here, a = 5 and a₄ = 9(1/2). To find: a₂ and a₃ Given that: a₄ = a + (4 -1)d = 9(1/2 ⇒ 5 + 3d = 19/2 ⇒ 3d = 19/2 - 5 = 9/2 ⇒ d = 3/2 Therefore, a₂ = a + d = 5 + 3/2 = 6(1/2) and a₃ = a + 2d = 5 + 2(3/2) = 8
Here, a = 5 and a₄ = 9(1/2). To find: a₂ and a₃
Given that: a₄ = a + (4 -1)d = 9(1/2
⇒ 5 + 3d = 19/2 ⇒ 3d = 19/2 – 5 = 9/2 ⇒ d = 3/2
Therefore, a₂ = a + d = 5 + 3/2 = 6(1/2) and a₃ = a + 2d = 5 + 2(3/2) = 8
Here, a₂ = 13 and a₄ = 3. To find: a₁ and a₃ Given that: a₂ = a + (2 -1)d = 13 ⇒ a + d = 13 ⇒ a = 13 - d ....(1) and a₄ = 3 ⇒ a + 3d = 3 Putting the value of a from equation (1), we get 13 - d + 3d = 3 ⇒ d = - 5 Putting the value of d in eqation (1), we get a = 13 - (-5) = 18 Therefore, a₁ = 18 andRead more
Here, a₂ = 13 and a₄ = 3. To find: a₁ and a₃
Given that: a₂ = a + (2 -1)d = 13
⇒ a + d = 13
⇒ a = 13 – d ….(1)
and a₄ = 3 ⇒ a + 3d = 3
Putting the value of a from equation (1), we get
13 – d + 3d = 3 ⇒ d = – 5
Putting the value of d in eqation (1), we get
a = 13 – (-5) = 18
Therefore, a₁ = 18 and a₃ = a + (3 – 1)d = 18 + 2(-5) = 8
In the following APs, find the missing terms in the boxes : [], 38, [], [], [], -22
In the following APs, find the missing terms in the boxes : [], 38, [], [], [], -22 Here, a₂ = 38 and a₆ = - 22. To find: a₁, a₃, a₄ and a₅ Given that: a₂ = a + (2 -1)d = 38 ⇒ a + d = 38 ⇒ a = 38 - d and a₆ = - 22 ⇒ a + 5d = - 22 Putting the value of a from equation (1), we get 38 - d + 5d = - 22 ⇒Read more
In the following APs, find the missing terms in the boxes : [], 38, [], [], [], -22
See lessHere, a₂ = 38 and a₆ = – 22. To find: a₁, a₃, a₄ and a₅
Given that: a₂ = a + (2 -1)d = 38
⇒ a + d = 38
⇒ a = 38 – d
and a₆ = – 22 ⇒ a + 5d = – 22
Putting the value of a from equation (1), we get
38 – d + 5d = – 22 ⇒ d = – 15
Putting the value of a from equation (1), we get
a = 38 – (-15) = 53
Therefore, a₁ = a = 53
a₃ = a + 2d = 53 +2(-15) = 23
a₄ = a + 3d = 53 + 3 (-15) = 8
a₅ = a + 4d = 53 + 4 (-15) = – 7
In the following APs, find the missing terms in the boxes : -4, [], [], [], [], 6
Here, a = - 4 and a₆ = 6. To find: a₂, a₃, a₄ and a₅ Given that: a₆ = a + (6 -1)d = 6 ⇒ -4 + 5d = 6 ⇒ 5d = 10 ⇒ d = 2 Therefore, a₂ = a + d = - 4 + 2 = - 2 a₃ = a + 2d = - 4 + 2(2) = 0 a₄ = a + 3d = - 4 + 3(2) = 2 a₅ = a + 4d = - 4 + 4(2) = 4
Here, a = – 4 and a₆ = 6. To find: a₂, a₃, a₄ and a₅
See lessGiven that: a₆ = a + (6 -1)d = 6
⇒ -4 + 5d = 6 ⇒ 5d = 10 ⇒ d = 2
Therefore, a₂ = a + d = – 4 + 2 = – 2
a₃ = a + 2d = – 4 + 2(2) = 0
a₄ = a + 3d = – 4 + 3(2) = 2
a₅ = a + 4d = – 4 + 4(2) = 4
In the following APs, find the missing terms in the boxes : 5, _, _, 9(1/2)
Here, a = 5 and a₄ = 9(1/2). To find: a₂ and a₃ Given that: a₄ = a + (4 -1)d = 9(1/2 ⇒ 5 + 3d = 19/2 ⇒ 3d = 19/2 - 5 = 9/2 ⇒ d = 3/2 Therefore, a₂ = a + d = 5 + 3/2 = 6(1/2) and a₃ = a + 2d = 5 + 2(3/2) = 8
Here, a = 5 and a₄ = 9(1/2). To find: a₂ and a₃
See lessGiven that: a₄ = a + (4 -1)d = 9(1/2
⇒ 5 + 3d = 19/2 ⇒ 3d = 19/2 – 5 = 9/2 ⇒ d = 3/2
Therefore, a₂ = a + d = 5 + 3/2 = 6(1/2) and a₃ = a + 2d = 5 + 2(3/2) = 8
In the following APs, find the missing terms in the boxes : _, 13, _, 3
Here, a₂ = 13 and a₄ = 3. To find: a₁ and a₃ Given that: a₂ = a + (2 -1)d = 13 ⇒ a + d = 13 ⇒ a = 13 - d ....(1) and a₄ = 3 ⇒ a + 3d = 3 Putting the value of a from equation (1), we get 13 - d + 3d = 3 ⇒ d = - 5 Putting the value of d in eqation (1), we get a = 13 - (-5) = 18 Therefore, a₁ = 18 andRead more
Here, a₂ = 13 and a₄ = 3. To find: a₁ and a₃
See lessGiven that: a₂ = a + (2 -1)d = 13
⇒ a + d = 13
⇒ a = 13 – d ….(1)
and a₄ = 3 ⇒ a + 3d = 3
Putting the value of a from equation (1), we get
13 – d + 3d = 3 ⇒ d = – 5
Putting the value of d in eqation (1), we get
a = 13 – (-5) = 18
Therefore, a₁ = 18 and a₃ = a + (3 – 1)d = 18 + 2(-5) = 8
In the following APs, find the missing terms in the boxes : 2, _ , 26
Here, a = 2 and a₃ = 26. To find: a₂ Given that: a₃ = a + (3 -1)d = 26 ⇒ 2 + 2d = 26 ⇒ d = 12 - d ....(1) and a₄ = 3 ⇒ a + 3d = 3 Therefore, a₂ = a + (2 - 1)d = 2 + 12 = 14
Here, a = 2 and a₃ = 26. To find: a₂
See lessGiven that: a₃ = a + (3 -1)d = 26
⇒ 2 + 2d = 26
⇒ d = 12 – d ….(1)
and a₄ = 3 ⇒ a + 3d = 3
Therefore, a₂ = a + (2 – 1)d = 2 + 12 = 14