If Tn-1 - 2Tn -12- n - and T10 - 192- then the 101th term of the sequence is

The correct option is D 55Given : nbsp;T_n+1 = 2T_n +12 ⇒ T_n+1 nbsp; T_n = 12 So, the given sequence is an A.P. with d = 12 Let the first term be a Now, ...

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Byju's Answer

Standard XI

Mathematics

Geometric Progression

If Tn+1 = 2Tn...

Question

If $T_{n + 1} = \frac{2 T_{n} + 1}{2}, n \in N$ and $T_{10} = \frac{19}{2}$ , then the $101^{t h}$ term of the sequence is

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Solution

The correct option is D $55$
Given : $T_{n + 1} = \frac{2 T_{n} + 1}{2}$
$\Rightarrow T_{n + 1} - T_{n} = \frac{1}{2}$
So, the given sequence is an A.P. with $d = \frac{1}{2}$
Let the first term be $a$
Now,
$T_{10} = \frac{19}{2} \Rightarrow a + 9 \times \frac{1}{2} = \frac{19}{2} \Rightarrow a = 5$
Therefore,
$T_{101} = 5 + 100 \times \frac{1}{2} = 55$

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If Tn+1=2Tn+12,n∈N and T10=192, then the 101th term of the sequence is

The correct option is D 55Given : Tn+1=2Tn+12 ⇒Tn+1−Tn=12 So, the given sequence is an A.P. with d=12 Let the first term be a Now, T10=192⇒a+9×12=192⇒a=5 Therefore, T101=5+100×12=55

If $T_{n + 1} = \frac{2 T_{n} + 1}{2}, n \in N$ and $T_{10} = \frac{19}{2}$ , then the $101^{t h}$ term of the sequence is