For a sequence an a1=2 and (n+1/an)=(1/3). Then displaystyle∑r=120ar is

Question

For a sequence an a1=2 and (n+1/an)=(1/3). Then displaystyle∑r=120ar is (A) (20/2)[4+19×3] (B) 3(1-(1/320)) (C) 2(1-320) (D) none of these.

Tardigrade Admin · Accepted Answer

(an+1/an) = (1/3) ⇒ Common ratio = (1/3) = R (say) First term = a1=2 ∑ limitsr=020 ar = (a1[1-R20]/1-R) = (2[1-((1/3))20]/1-(1)3) = (2[1-(1/320)]/(2)3) = 3[1-(1/320)]

Q. For a sequence ${a_{n}}, a_{1} = 2$ and $\frac{n + 1}{a _{n}} = \frac{1}{3}$ . Then $r = 1 \sum 20 a_{r}$ is

$\frac{20}{2} [4 + 19 \times 3]$

$3 (1 - \frac{1}{3 ^{20}})$

$2 (1 - 3^{20})$

none of these.

Q. For a sequence {an​},a1​=2 and an​n+1​=31​. Then r=1∑20​ar​ is

220​[4+19×3]

3(1−3201​)

2(1−320)

none of these.

an​an+1​​=31​ ⇒ Common ratio =31​=R (say) First term =a1​=2 r=0∑20​ar​=1−Ra1​[1−R20]​=1−31​2[1−(31​)20]​ =32​2[1−3201​]​=3[1−3201​]

Q. For a sequence ${a_{n}}, a_{1} = 2$ and $\frac{n + 1}{a _{n}} = \frac{1}{3}$ . Then $r = 1 \sum 20 a_{r}$ is

$\frac{20}{2} [4 + 19 \times 3]$

$3 (1 - \frac{1}{3 ^{20}})$

$2 (1 - 3^{20})$