Question

Verify that following is an $AP$, and then write the sum of its next three terms. $a,2a+1,3a+2,4a+3,...$

Answer 1

We have
$(2a+1)-a=a+1$
$(3a+2)-(2a+1)=a+1$
This shows that difference of a term and the preceding term is always same.Hence, the given sequence form an AP.
Here first term $A=a$ and common difference $D=a+1$
Therefore,
$A_5=A+4D=a+4\times (a+1)$
$\Rightarrow A_5=5a+4$
$A_6=A+5D=a+5\times (a+1)$
$\Rightarrow A_6=6a+5$
$A_7=A+6D=a+6\times (a+1)$
$\Rightarrow A_7=7a+6$
Hence, the next three terms are
$5a+4,6a+5,7a+6$
Sum=$5a+4+6a+5+7a+6$=33