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
Here, a₁₁ = 38 and a₁₆ = 73. To find: a₃₁ Given that: a₁₁ = a + (11 - 1)d = 38 ⇒ a + 10d = 38 ⇒ a = 38 - 10d ....(1) and a₁₆ = 73 ⇒ a + 15d = 73 Putting the value of a from equation (1), we get 38 - 10d + 15d = 73 ⇒ 5d = 35 ⇒ d = 7 Putting the value of d in equation (1), we get a = 38 - 10(7) = - 32Read more
Here, a₁₁ = 38 and a₁₆ = 73. To find: a₃₁
Given that: a₁₁ = a + (11 – 1)d = 38
⇒ a + 10d = 38
⇒ a = 38 – 10d ….(1)
and a₁₆ = 73 ⇒ a + 15d = 73
Putting the value of a from equation (1), we get
38 – 10d + 15d = 73
⇒ 5d = 35 ⇒ d = 7
Putting the value of d in equation (1), we get
a = 38 – 10(7) = – 32
Therefore, a₃₁ = a + 30d = – 32 + 30(7) = 178
Hence, the 31st term is 178.
Here, a = 11 and d = 8 - 11 = - 3. Let, the nth term of the A.P. is - 150. Therefore, a_n = - 150 ⇒ a + (n - 1)d = - 150 ⇒ 11 + (n - 1)(- 3) = - 150 ⇒ 11 - 3n + 3 = - 150 ⇒ - 3n = - 164 ⇒ n = 164/3 = 54(2/3) Here, n is not a natural number, therefore, - 150 is not the term of A.P., 11, 8, 5, 2 . . .
Here, a = 11 and d = 8 – 11 = – 3.
Let, the nth term of the A.P. is – 150.
Therefore, a_n = – 150
⇒ a + (n – 1)d = – 150
⇒ 11 + (n – 1)(- 3) = – 150
⇒ 11 – 3n + 3 = – 150
⇒ – 3n = – 164
⇒ n = 164/3 = 54(2/3)
Here, n is not a natural number, therefore, – 150 is not the term of A.P., 11, 8, 5, 2 . . .
Here, a = 7 and d = 13 - 7 = 6. Let, the total number of terms in the A.P. is n. Therefore, a_n = 205 ⇒ a + (n - 1)d = 205 ⇒ 7 + (n - 1)(6) = 205 ⇒ (n - 1)(6) = 198 ⇒ n - 1 = 33 ⇒ n = 34 Hence, there are 34 terms in the given A.P.
Here, a = 7 and d = 13 – 7 = 6.
Let, the total number of terms in the A.P. is n.
Therefore, a_n = 205
⇒ a + (n – 1)d = 205
⇒ 7 + (n – 1)(6) = 205
⇒ (n – 1)(6) = 198
⇒ n – 1 = 33 ⇒ n = 34
Hence, there are 34 terms in the given A.P.
Here, a = 18 and d = 15(1/2) - 18 = - 5/2. Let, the total number of terms in the A.P. is n. Therefore, a_n = - 47 ⇒ a + (n - 1)d = - 47 ⇒ 18 + (n - 1)(- 5/2) = - 47 ⇒ (n - 1)(- 5/2) = - 65 ⇒ n - 1 = 26 ⇒ n = 27 Hence, there are 27 terms in the given A.P.
Here, a = 18 and d = 15(1/2) – 18 = – 5/2.
Let, the total number of terms in the A.P. is n.
Therefore, a_n = – 47
⇒ a + (n – 1)d = – 47
⇒ 18 + (n – 1)(- 5/2) = – 47
⇒ (n – 1)(- 5/2) = – 65
⇒ n – 1 = 26
⇒ n = 27
Hence, there are 27 terms in the given A.P.
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
Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
Here, a₁₁ = 38 and a₁₆ = 73. To find: a₃₁ Given that: a₁₁ = a + (11 - 1)d = 38 ⇒ a + 10d = 38 ⇒ a = 38 - 10d ....(1) and a₁₆ = 73 ⇒ a + 15d = 73 Putting the value of a from equation (1), we get 38 - 10d + 15d = 73 ⇒ 5d = 35 ⇒ d = 7 Putting the value of d in equation (1), we get a = 38 - 10(7) = - 32Read more
Here, a₁₁ = 38 and a₁₆ = 73. To find: a₃₁
See lessGiven that: a₁₁ = a + (11 – 1)d = 38
⇒ a + 10d = 38
⇒ a = 38 – 10d ….(1)
and a₁₆ = 73 ⇒ a + 15d = 73
Putting the value of a from equation (1), we get
38 – 10d + 15d = 73
⇒ 5d = 35 ⇒ d = 7
Putting the value of d in equation (1), we get
a = 38 – 10(7) = – 32
Therefore, a₃₁ = a + 30d = – 32 + 30(7) = 178
Hence, the 31st term is 178.
Check whether – 150 is a term of the AP : 11, 8, 5, 2 . . .
Here, a = 11 and d = 8 - 11 = - 3. Let, the nth term of the A.P. is - 150. Therefore, a_n = - 150 ⇒ a + (n - 1)d = - 150 ⇒ 11 + (n - 1)(- 3) = - 150 ⇒ 11 - 3n + 3 = - 150 ⇒ - 3n = - 164 ⇒ n = 164/3 = 54(2/3) Here, n is not a natural number, therefore, - 150 is not the term of A.P., 11, 8, 5, 2 . . .
Here, a = 11 and d = 8 – 11 = – 3.
See lessLet, the nth term of the A.P. is – 150.
Therefore, a_n = – 150
⇒ a + (n – 1)d = – 150
⇒ 11 + (n – 1)(- 3) = – 150
⇒ 11 – 3n + 3 = – 150
⇒ – 3n = – 164
⇒ n = 164/3 = 54(2/3)
Here, n is not a natural number, therefore, – 150 is not the term of A.P., 11, 8, 5, 2 . . .
Find the number of terms in each of the following APs : 7, 13, 19, . . . , 205
Here, a = 7 and d = 13 - 7 = 6. Let, the total number of terms in the A.P. is n. Therefore, a_n = 205 ⇒ a + (n - 1)d = 205 ⇒ 7 + (n - 1)(6) = 205 ⇒ (n - 1)(6) = 198 ⇒ n - 1 = 33 ⇒ n = 34 Hence, there are 34 terms in the given A.P.
Here, a = 7 and d = 13 – 7 = 6.
See lessLet, the total number of terms in the A.P. is n.
Therefore, a_n = 205
⇒ a + (n – 1)d = 205
⇒ 7 + (n – 1)(6) = 205
⇒ (n – 1)(6) = 198
⇒ n – 1 = 33 ⇒ n = 34
Hence, there are 34 terms in the given A.P.
Find the number of terms in each of the following APs : 18, 15(1/2), 13, . . . , – 47
Here, a = 18 and d = 15(1/2) - 18 = - 5/2. Let, the total number of terms in the A.P. is n. Therefore, a_n = - 47 ⇒ a + (n - 1)d = - 47 ⇒ 18 + (n - 1)(- 5/2) = - 47 ⇒ (n - 1)(- 5/2) = - 65 ⇒ n - 1 = 26 ⇒ n = 27 Hence, there are 27 terms in the given A.P.
Here, a = 18 and d = 15(1/2) – 18 = – 5/2.
See lessLet, the total number of terms in the A.P. is n.
Therefore, a_n = – 47
⇒ a + (n – 1)d = – 47
⇒ 18 + (n – 1)(- 5/2) = – 47
⇒ (n – 1)(- 5/2) = – 65
⇒ n – 1 = 26
⇒ n = 27
Hence, there are 27 terms in the given A.P.