Given a sequence of 4 numbers, first three of which are in G.P. and the last three are in A.P. with common difference 6. If first and last terms of this sequence are equal, then the last term is

Question

Given a sequence of 4 numbers, first three of which are in G.P. and the last three are in A.P. with common difference 6. If first and last terms of this sequence are equal, then the last term is (A) 2 (B) 4 (C) 8 (D) 16

Tardigrade Admin · Accepted Answer

Let a, ar, a2, a. Now, 2 a2=a r+a <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/bd594030c255c850ca84795887fa6e8d-.png" /> As, a r-a r2=6 ⇒ a=8

Q. Given a sequence of 4 numbers, first three of which are in G.P. and the last three are in A.P. with common difference 6. If first and last terms of this sequence are equal, then the last term is

2

4

8

16

Let a, ar, $a^{2}$ , a.
Now, $2 a^{2} = a r + a$

As, $a r - a r^{2} = 6 \Rightarrow a = 8$

Q. Given a sequence of 4 numbers, first three of which are in G.P. and the last three are in A.P. with common difference 6. If first and last terms of this sequence are equal, then the last term is

2

4

8

16

Let a, ar, a2, a. Now, 2a2=ar+a As, ar−ar2=6⇒a=8

Let a, ar, $a^{2}$ , a.
Now, $2 a^{2} = a r + a$

As, $a r - a r^{2} = 6 \Rightarrow a = 8$