(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,
$\mathrm{S}_{\mathrm{n}}=\mathrm{n} / 2[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\mathrm{S}_{12}=12 / 2[2(-37)+(12-1) \times 4]$
$=6[-74+11 \times 4]$
$=6[-74+44]$
$=6(-30)=-180$