Question

Verify that following is an AP, and then write its next three terms.

a + b, (a + 1) + b, (a + 1) + (b + 1), ...

Answer 1

We have
$\left(\right(a+1)+b)-(a+b)=1$
$\left(\right(a+1)+(b+1\left)\right)-\left(\right(a+1)+b)=1$
This shows that difference of a term and the preceding term is always same.Hence, the given sequence form an AP.

nth term $t_n=a+(n-1)d$

$a=$ First term

$d=$ Common difference

$n=$ number of terms

$t_n=n^{th}$ term

Here first term $A=a+b$ and common difference $D=1$
Therefore,
$A_4=A+3D=(a+b)+3\times 1=a+b+2+1$
$\Rightarrow A_4=(a+2)+(b+1)$

$A_5=A+4D=(a+b)+4\times 1=a+b+2+2$
$\Rightarrow A_5=(a+2)+(b+2)$

$A_6=A+5D=(a+b)+5\times 1=a+b+3+2$
$\Rightarrow A_6=(a+3)+(b+2)$

Hence, the next three terms are
$(a+2)+(b+1),(a+2)+(b+2),(a+3)+(b+2)$