Which term of the A.P. 121 , 117 , 113 , … is its first negative term?

Here, A.P is 121 , 117 , 113 . . . . . We need to find first negative term of the A.P. a = 121 Now, Common difference d = a 1 - a = 117 - 1 21 = - 4 Now, we need to find the first negative term, a n < 0 121 + n - 1 - 4 < 0 121 - 4 n + 4 < 0 125 - 4 n < 0 4 n > 125 Further simplifying, we get, n > 125 4 n > 31 1 4 n ≥ 3 2 (as n is an integer) Thus, n = 32 Therefore, the first negative term is the 32 nd term of the given A.P.

Which term of the A.P. 121 , 117 , 113 , … is its first negative term?

Here, A.P is 121 , 117 , 113 . . . . . We need to find first negative term of the A.P. a = 121 Now, Common difference d = a 1 - a = 117 - 1 21 = - 4 Now, we need to find the first negative term, a n < 0 121 + n - 1 - 4 < 0 121 - 4 n + 4 < 0 125 - 4 n < 0 4 n > 125 Further simplifying, we get, n > 125 4 n > 31 1 4 n ≥ 3 2 (as n is an integer) Thus, n = 32 Therefore, the first negative term is the 32 nd term of the given A.P.

How many terms are there in the A.P. ? - 1 , - 5 6 , - 2 3 , - 1 2 , … , 10 3 .

Here, A.P. is - 1 , - 5 6 , - 2 3 , - 1 2 , … . , 10 3 . The first term ( a 1 ) = - 1 The last term a n = 10 3 Now, Common difference ( d ) = a 2 - a 1 = - 5 6 - ( - 1 ) = - 5 + 6 6 = 1 6 Thus, using the formula for n th term of A.P. a n = a + ( n - 1 ) d , we get 10 3 = - 1 + ( n - 1 ) 1 6 ⇒ 10 3 + 1 = 1 6 n - 1 6 ⇒ 13 3 + 1 6 = 1 6 n Further, solving for n , we get ⇒ 26 + 1 6 = 1 6 n ⇒ n = 27 6 ( 6 ) ∴ n = 27 Therefore, the number of terms present in the given A.P. is 27 .

How many terms are there in the A.P.? 7 , 13 , 19 , … , 205

Given A.P is 7 , 13 , 19 , … . 205 . The first term ( a 1 ) = 7 The last term a n = 205 Now, Common difference ( d ) = a 2 - a 1 = 13 - 7 = 6 Thus, using the formula for n th term of an A.P. a n = a + ( n - 1 ) d , we get : 205 = 7 + ( n - 1 ) 6 205 - 7 = 6 n - 6 198 + 6 = 6 n n = 204 6 n = 34 Thus, n = 34 . Therefore, the number of terms present in the given A.P is 34 .

How many terms are there in the A.P. ? 18 , 15 1 2 , 13 , … . , - 47 .

Here, A.P is 18 , 15 1 2 , 13 , … . , - 47 . The first term ( a ) = 18 The last term a n = - 47 Now, Common difference d = a 1 - a = 15 1 2 - 18 = 31 2 - 1 8 = 31 - 36 2 = - 5 2 We know the formula for the n th term of an A.P., a n = a + ( n - 1 ) d So, - 47 = 18 + n - 1 - 5 2 ⇒ - 47 - 18 = - 5 2 n + 5 2 ⇒ - 65 - 5 2 = - 5 2 n Further, solving for n , we get : ⇒ - 130 - 5 2 = - 5 2 n ⇒ - 135 2 ( 2 ) = - 5 n n = - 135 - 5 n = 27 Therefore, the number of terms present in the given A.P is 27 .