Question

Find the sum of each of the following APs:
(i) 2, 7, 12, 17,... to 19 terms.
(ii) 9, 7, 5, 3, ... to 14 terms.
(iii) -37, -33, -29, ... to 12 terms.
(iv) $\frac{1}{15},\frac{1}{12},\frac{1}{10},...to11$
(v) 0.6, 1.7, 2.8, ... to 100 terms.

Answer 1

Sum of n terms $S_n$= $\frac{n}{2}$ ($2a+(n-1)d$)
(i)
a = 2 , d = 7 - 2 = 5 and n = 19

$S_{19}$= $\frac{19}{2}$ ($2\times 2+(19-1)5$)

$S_{19}$= $\frac{19}{2}$ ($4+90$)

$S_{19}$= $\frac{19}{2}$ × 94

$S_{19}$ = 893

(ii)
9, 7, 5, 3, ... to 14 terms.
a=9, d=-2, n=14
$S_{14}$ = $\frac{14}{2}${2 × 9+(14-1)×-2}
$S_{14}$ = 7{18-26}
$S_{14}$ = 7×-8
$S_{14}$ = -56

(iii)
a = -37 , d = 4 and n = 12
$S_{12}$ = $\frac{12}{2}${ -37×2 + (12-1)×4}
$S_{12}$ = 6{ -74 + 44}
$S_{12}$ = -180

(iv)

a = $\frac{1}{15}$ , d = $\frac{1}{12}$ - $\frac{1}{15}$ = $\frac{1}{60}$
$S_{11}$ = $\frac{11}{2}$ { $\frac{2}{15}$ + 10 × $\frac{1}{60}$ }
$S_{11}$ = $\frac{11}{2}$ { $\frac{18}{60}$ }
$S_{11}$ = $\frac{33}{20}$

(v)

a = 0.6 , d = 1.1 and n = 100
$S_{100}$ = $\frac{100}{2}${ 2 × 0.6 + (100-1)×1.1 }
$S_{100}$ = 50 × { 1.2 + 108.9}
$S_{100}$ = 50 × 110.1
$S_{100}$ = 5505