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 = 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.
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 : 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.