The sum of first three terms of a GP is 39 10 and their product is 1 . Find the common ratio and these three terms.

Let the first three terms of G.P. be a r , a , a r It is given that a r × a × a r = 1 ⇒ a 3 = 1 ⇒ a = 1 And a r + a + a r = 39 10 ⇒ a 1 r + 1 + r = 39 10 ⇒ 1 r + 1 + r = 39 10 … ( a = 1 ) ⇒ 1 r + r = 39 10 - 1 = 29 10 ⇒ 10 1 + r 2 = 29 r ⇒ 10 r 2 - 29 r + 10 = 0 ⇒ 10 r 2 - 25 r - 4 r + 10 = 0 ⇒ 5 r ( 2 r - 5 ) - 2 ( 2 r - 5 ) = 0 ⇒ ( 2 r - 5 ) ( 5 r - 2 ) = 0 ⇒ r = 5 2 · 2 5 Therefore the first three terms are: i ) If r = 5 2 then 2 5 , 1 , 5 2 ii ) If r = 2 5 then the terms are 5 2 , 1 , 2 5

The sum of first three terms of a GP is 39 10 and their product is 1 . Find the common ratio and these three terms.

Let the first three terms of G.P. be a r , a , a r It is given that a r × a × a r = 1 ⇒ a 3 = 1 ⇒ a = 1 And a r + a + a r = 39 10 ⇒ a 1 r + 1 + r = 39 10 ⇒ 1 r + 1 + r = 39 10 … ( a = 1 ) ⇒ 1 r + r = 39 10 - 1 = 29 10 ⇒ 10 1 + r 2 = 29 r ⇒ 10 r 2 - 29 r + 10 = 0 ⇒ 10 r 2 - 25 r - 4 r + 10 = 0 ⇒ 5 r ( 2 r - 5 ) - 2 ( 2 r - 5 ) = 0 ⇒ ( 2 r - 5 ) ( 5 r - 2 ) = 0 ⇒ r = 5 2 · 2 5 Therefore the first three terms are: i ) If r = 5 2 then 2 5 , 1 , 5 2 ii ) If r = 2 5 then the terms are 5 2 , 1 , 2 5

Find the 6 t h and n t h terms of the GP 2 , 6 , 18 , 54 …

Given: GP is 2 , 6 , 18 , 54 … The given GP is of the form, a , a r , a r 2 , a r 3 … Where r is the common ratio. First term in the given GP a 1 = a = 2 Second term in GP a 2 = 6 Now, the common ratio, r = a 2 a 1 r = 6 2 = 3 We know that, n t h term of G P is, a n = a r n - 1 So, the 6 t h term in the G P , a 6 = a r 5 = 2 × 3 5 = 486 n t h term in the G P , a n = a r n - 1 = 2 . 3 n - 1 Hence, 6 t h term = 486 and n t h term = 2 . 3 n - 1 .

Find the 17 t h and n t h terms of the G P 2 , 2 2 , 4 , 8 2 …

Given G P is 2 , 2 2 , 4 , 8 2 … … The G P is of the form, a , a r , a r , 2 a r 3 … . Where r is the common ratio. First term in the given G P , a 1 = a = 2 Second term in G P , a 2 = 2 2 Now, the common ratio, r = a 2 a 1 r = 2 2 2 = 2 We know that, n t h term of G P is, a n = a r n - 1 So, the 17 t h term in the G P , a 17 = a r 16 = 2 × ( 2 ) 16 = 512 n t h term in the G P , a n = a r n - 1 = 2 ( 2 ) n - 1 = ( 2 ) n + 1 Hence, 17 t h term = 512 and n t h term = ( 2 ) n + 1 .

Find the 10 t h and n t h terms of the G P - 3 4 , 1 2 , - 1 3 , 2 9 …

Given G P is - 3 4 , 1 2 , - 1 3 , 2 9 … The G P is of the form, a , a r , a r 2 , a r 3 … . Where r is the common ratio. The first term in the given G P , a = a 1 = - 3 4 The second term in G P , a 2 = 1 2 Now, the common ratio, r = a 2 a 1 r = - 1 2 3 4 = - 2 3 We know that, n t h term of G P is, a n = ar n - 1 So, the 10 t h term, a 10 = a r 9 a 10 = a r 9 = - 3 4 - 2 3 9 = 128 6561 Now, the required n t h term, a n = a r n - 1 a n = - 3 4 - 2 3 n - 1 = 9 8 - 2 3 n Hence, the 10 t h term, a 10 = 128 6561 and n t h term, a n = 9 8 - 2 3 n

Which term of the GP 3 , 3 , 3 3 … is 729 ?

Given GP is 3 , 3 , 3 3 … The given GP is of the form, a , a r , a r 2 , a r 3 … . Where r is the common ratio. First term in the given GP , a 1 = a = 3 Second term in GP , a 2 = 3 Now, the common ratio, r = d 2 a 1 r = 3 3 = 3 Let us consider 729 as the n t h term of the GP . Now, n t h term of GP is, a n = a r n - 1 729 = 3 ( 3 ) n - 1 3 n = 3 12 n = 12 So, 729 is the 12 t h term in GP .