Question

How many terms are there in the series:
(i) $4,7,10,13,...........................................,148$.
(ii) $0.5,0.53,0.56,...................................,1.1$.
(iii) $\frac{3}{4},1,1\frac{1}{4},...............................,3$.

Answer 1

$4,7,10,13,.........,148$

$a=4,d=-4=3$

$T_n=a+(n-1)d$

$148=4+(n-1)3$

$144=3n-3$

$n=\frac{147}{3}$

$\therefore n=49$

49th term is (i)

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$0.5,0.53,0.56,.........1.1$

$a=0.5,d=0.53-0.5=0.03$

$T_n=a+(n-1)d$

$1.1=0.5+(n-1)0.03$

$0.6=0.03n-0.03$

$n=\frac{0.63}{0.03}$

$\therefore n=21$

21th term is (ii)

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$\frac{3}{4},1,1\frac{1}{4},............,3$

$a=\frac{3}{4},d=1-\frac{3}{4}=\frac{1}{4}$

$T_n=a+(n-1)d$

$3=\frac{3}{4}+(n-1)\frac{1}{4}$

$\frac{9}{4}=\frac{n}{4}-\frac{1}{4}$

$\frac{10}{4}=\frac{n}{4}$

$\therefore n=10$

10th term is (iii)