Question

Verify the $AP:5,\frac{14}{3},\frac{13}{3},\mathrm{4....}$ and find the sum of the next three terms:

Answer 1

We have,
$\frac{14}{3}-5=\frac{-1}{3};$

$\frac{13}{3}-\frac{14}{3}=\frac{-1}{3};$
This shows that difference of a term and the preceding term is always same.Hence, the given sequence form an AP.
Here first term $a=5$ and common difference $d=\frac{-1}{3}$
Therefore,
$a_5=a+4d=5+4\times \frac{-1}{3}$
$\Rightarrow a_5=\frac{11}{3}$
$a_6=a+5d=5+5\times \frac{-1}{3}$
$\Rightarrow a_5=\frac{10}{3}$
$a_7=a+6d=5+6\times \frac{-1}{3}$
$\Rightarrow a_7=3$
Hence, the next three terms are

$\frac{11}{3},\frac{10}{3},3$