Which term of the G.P.: 2 , 2 2 , 4 , . . . is 128 ?

Given progression is 2 , 2 2 , 4 , … … … . 128 The terms of the G.P. are in the form of a , a r , a r 2 , a r 3 , . . . a r n - 1 are called as a 1 , a 2 , a 3 , a 4 , . . . , a n respectively. Common ratio r = a 2 a 1 = a 3 a 2 = a n a n - 1 a 1 = 2 a 2 = 2 2 a n = 128 Common Ratio r = a 2 a 1 = 2 2 2 = 2 a n = a 1 r n - 1 ⇒ 128 = 2 2 n - 1 ⇒ 64 = 2 n - 1 2 ⇒ n - 1 2 = 6 ⇒ n = 13 So,it is 13 th term

Which term of G.P. 1 , 1 3 , 1 9 , . . . . . is 1 243 ?

Given progression is 1 , 1 3 , 1 9 , . . . . . The terms of the G.P. are in the form of a , a r , a r 2 , a r 3 , . . . a r n - 1 are called as a 1 , a 2 , a 3 , a 4 , . . . , a n respectively. Common ratio r = a 2 a 1 = a 3 a 2 = a n a n - 1 Here, a 1 = 1 a 2 = 1 3 a n = 1 243 Common Ratio r = a 2 a 1 = 1 3 1 = 1 3 a n = a 1 r n - 1 ⇒ 1 243 = 1 1 3 n - 1 ⇒ 5 = n - 1 ⇒ n = 6 So, it is 6 th term.

Find the number of terms of a G.P. whose first term is 3 4 and common ratio is 2 and the last term is 384 .

Given a = 3 4 and r = 2 and t n = 384 The n t h term of an G.P. is given by t n = a r n - 1 Now, 384 = 3 4 ( 2 ) n - 1 512 = ( 2 ) n - 1 ( 2 ) 9 = ( 2 ) n - 1 9 = n – 1 10 = n Therefore , number of terms are 10

For what values of x , the numbers - 2 7 , x , - 7 2 are in G.P.?

Given that, Numbers are - 2 7 , x , - 7 2 If three numbers a , b and c are in G.P. then, b a = c b Therefore, x - 2 7 = - 7 2 x ⇒ - 7 x 2 = - 7 2 x ⇒ x 2 = - 2 × 7 - 2 × 7 = 1 ⇒ x = ± 1 = ± 1 Therefore, for x = ± 1 the given numbers are in G.P.

The first term of an A.P is 5 , the last term is 45 and the sum is 400 . Find the number of terms and the common difference.

Given, a = 5 . Last Term of an A.P l = 45 . Sum of n terms, S n = 400 . We know that: S n = n 2 a + l ⇒ 400 = n 2 5 + 45 n = 800 50 = 16 Hence, there are 16 terms in the given A.P. We know that general terms of an arithmetic progression as: a n = a + n - 1 d a 16 = a + 15 d 45 = 5 + 15 d d = 45 15 = 8 3 Hence, the common difference of an A.P is 8 3 .

Solve: 1 + 4 + 7 + 10 + . . . + x = 287

1 + 4 + 7 + 10 + … + x = 287 Here a = 1 ; d = 3 & T n = x ∵ T n = x ⇒ a + n - 1 d = x ⇒ 1 + n - 1 × 3 = x ⇒ 3 n - 2 = x ⇒ n = x + 2 3 … ( i ) S n = n 2 a + l ⇒ 287 = n 2 1 + x ⇒ 574 = x + 2 3 ( 1 + x ) from – ( i ) ⇒ 1722 = x 2 + 3 x + 2 ⇒ x 2 + 3 x - 1720 = 0 ⇒ x + 43 x - 40 = 0 ⇒ x = - 43 , & x = 40 ∵ Sequence is increasing ∴ x = - 43 (reject) ∴ x = 40