A) \[99\]
B) \[101\]
C) \[103\]
D) \[191\]
Correct Answer: B
Solution :
Given an AP of odd numbers of terms n. \[d=\frac{1}{10}\] Let \[{{T}_{1}}=(2k+1)\] is odd. \[{{T}_{n}}=a+(n-1)d\] \[=(2k+1)+(n-1)\left( \frac{1}{10} \right)\] \[=2k+1+\frac{n}{10}-\frac{1}{10}\] \[\Rightarrow \] \[{{T}_{n}}=2k+\frac{n}{10}+\frac{9}{10}\] According to the question, \[\Rightarrow \] \[{{T}_{1}}+{{T}_{n}}=10\] \[(2k+1)+\left( 2k+\frac{n}{10}+\frac{9}{10} \right)=10\] ?.(i) and \[{{T}_{1}}\,\,.\,\,\,{{T}_{n}}=0\] \[\Rightarrow \] \[(2k+1)\left( 2k+\frac{n}{10}+\frac{9}{10} \right)=0\] ?..(ii) If \[2k+1=0\] \[\Rightarrow \] \[k=-\frac{1}{2}\] Then, from Eq. (i), we get \[0+\left( 2\times \frac{(-1)}{2}+\frac{n}{10}+\frac{9}{10} \right)=10\] \[\Rightarrow \] \[-1+\frac{n}{10}+\frac{9}{10}=10\] \[\Rightarrow \] \[\frac{n}{10}-\frac{1}{10}=10\] \[\Rightarrow \] \[\frac{n-1}{10}=10\] \[\Rightarrow \] \[n-1=100\] \[\Rightarrow \] \[n=101\] If \[2k+\frac{n}{10}+\frac{9}{10}=0\] \[\Rightarrow \] \[2k=-\left( \frac{n}{10}+\frac{9}{10} \right)\] Then, from Eq. (i), we get \[\left[ -\left( \frac{n}{10}+\frac{9}{10} \right)+1 \right]+\left[ -\left( \frac{n}{10}+\frac{9}{10} \right)+\frac{n}{10}+\frac{9}{10} \right]=10\] \[\Rightarrow \] \[\frac{-n}{10}-\frac{9}{10}+1=10\] \[\Rightarrow \] \[\frac{-n}{10}+\frac{1}{10}=10\] \[\Rightarrow \] \[-n+1=100\] \[\Rightarrow \] \[-n=99\] \[\Rightarrow \] \[n=-99,\] which is not possible. Hence, number of terms \[n=101\] \[[\because \,\,\,n\ne -ve]\]
You need to login to perform this action.
You will be redirected in
3 sec
