(i) $2,7,12, \ldots .$, to 10 terms.

(ii) $-37,-33,-29, \ldots$, to 12 terms

Solutions:

(i) Provided that, $2,7,12, \ldots$, to 10 terms

And for this A.P.,

The first term, $\mathrm{a}=2$

And the common difference, $d=a_{2}-a_{1}=7-2=5$

$n=10$

We all know that,for sum of the nth term in the AP series the formula is,

$S_{n}=n / 2[2 a+(n-1) d]$

$S_{10}=10 / 2[2(2)+(10-1) \times 5]$

$=5[4+(9) \times(5)]$

$=5 \times 49=245$

(ii) Provided that, $-37,-33,-29, \ldots$, to 12 terms

And for this A.P.,

first term, $\mathrm{a}=-37$

And the common difference, $d=a_{2}-a_{1}$

$d=(-33)-(-37)$

$=-33+37=4$

$n=12$

We all know that, for sum of the nth term in the AP series the formula is,

$\mathrm{S}_{\mathrm{n}}=\mathrm{n} / 2[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$

$\mathrm{S}_{12}=12 / 2[2(-37)+(12-1) \times 4]$

$=6[-74+11 \times 4]$

$=6[-74+44]$

$=6(-30)=-180$