Question
Write $5^{th}$ term from the end of the A.P. 3$, 5, 7, 9, ....., 201$
✓
Answer
Given,
A.P, $3, 5, 7, 9, ..... 201$
Here, First term $a=3$
Difference $d =5-3=2$
and Last term $a _{ n }=201$
We knaw,
$a_n=a+(n-1) d$
$\Rightarrow 201=3+(n-1) 2$
$\Rightarrow 201=3+2 n-2$
$\Rightarrow 201=1+2 n$
$\Rightarrow 2 n=201-1$
$\Rightarrow 2 n=200$
$\Rightarrow n=\frac{100}{2}$
$\Rightarrow n=100$
Now, we have to find $5^{\text {th }}$ term from the end $100^{\text {th }}-4^{\text {th }}=96^{\text {th }}$
$a_n=a+(n-1) d$
$\Rightarrow a_{96}=3+(96-1) 2$
$\Rightarrow a_{96}=3+95 \times 2$
$\Rightarrow a_{96}=3+190$
$\Rightarrow a_{96}=193$
Hence, $5^{\text {th }}$ term from the end of given A.P. is $193.$
A.P, $3, 5, 7, 9, ..... 201$
Here, First term $a=3$
Difference $d =5-3=2$
and Last term $a _{ n }=201$
We knaw,
$a_n=a+(n-1) d$
$\Rightarrow 201=3+(n-1) 2$
$\Rightarrow 201=3+2 n-2$
$\Rightarrow 201=1+2 n$
$\Rightarrow 2 n=201-1$
$\Rightarrow 2 n=200$
$\Rightarrow n=\frac{100}{2}$
$\Rightarrow n=100$
Now, we have to find $5^{\text {th }}$ term from the end $100^{\text {th }}-4^{\text {th }}=96^{\text {th }}$
$a_n=a+(n-1) d$
$\Rightarrow a_{96}=3+(96-1) 2$
$\Rightarrow a_{96}=3+95 \times 2$
$\Rightarrow a_{96}=3+190$
$\Rightarrow a_{96}=193$
Hence, $5^{\text {th }}$ term from the end of given A.P. is $193.$
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