Find the 37th term of the AP

Question:

(i) Find the 37 th term of the AP $6,7 \frac{3}{4}, 9 \frac{1}{2}, 11 \frac{1}{4}, \ldots$

(ii) Find the 25 th term of the AP $5,4 \frac{1}{2}, 4,3 \frac{1}{2}, 3, \ldots$

 

Solution:

(i) The given AP is $6,7 \frac{3}{4}, 9 \frac{1}{2}, 11 \frac{1}{4}, \ldots$

First term, $a=6$ and common difference, $d=7 \frac{3}{4}-6 \Rightarrow \frac{31}{4}-6 \Rightarrow \frac{31-24}{4}=\frac{7}{4}$

Now, $T_{37}=a+(37-1) d=a+36 d$

$=6+36 \times \frac{7}{4}=6+63=69$

$\therefore 37^{t h}$ term $=69$

(ii) The given AP is $5,4 \frac{1}{2}, 4,3 \frac{1}{2}, 3, \ldots$

 First term = 5

Common difference $=4 \frac{1}{2}-5 \Rightarrow \frac{9}{2}-5 \Rightarrow \frac{9-10}{2}=-\frac{1}{2}$

$\therefore a=5$ and $d=-\frac{1}{2}$

Now, $T_{25}=a+(25-1) d=a+24 d$

$=5+24 \times\left(-\frac{1}{2}\right)=5-12=-7$

$\therefore 25^{\text {th }}$ term $=-7$

 

 

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