If a

_{1}= 1 and a_{n+1}= 2a_{n}+5, n=1,2,… then a_{100}is equal to
(1)(5*2

Solution follows here:^{99}- 6) (2) (5*2^{99}+ 6) (3) (6*2^{99}+ 5) (4)(6*2^{99}- 5)
Solution:

The better way is to find some values of a

_{n}and check with the given answer options:
Before proceeding further, let us convert the given answer
options in terms of n:

a

_{n}=
(1)(5*2

^{n-1}– 6) (2)(5*2^{n-1}+ 6) (3)(6*2^{n-1}+ 5) (4)(6*2^{n-1}– 5)
Given a

_{1}= 1 this is true only for option (4).
a1 = 6*2

^{1-1}– 5 = 6*2^{0}– 5 = 6-5 =1
No other option gives us value 1. Hence answer is option(4).

But for our satisfaction let us try for some more values of a

_{n}:
a

_{n+1}= 2a_{n}+5 => a_{2}= 2a_{1}+5 = 2+5 = 7; substitute n = 2 in option (4) => a_{2}= 6*2^{1}-5 = 7
yes, verified correct.

a

_{3}= 2a_{2}+5 = 2(7)+5 = 19; substitute n = 3 in option (4) => a_{3}= 6*2^{2}-5 = 24-5 = 19
its true, verified correct.

Answer (4)
Hey great thanks a ton pls upload on CAT 2013, 12 n 11 question types

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