Question

Find the sum of the following APs:

1. $2, 7, 12, .......,$ to $10$ terms.
2. $–37, –33, –29,.......,$ to $12$ terms.
3. $0.6, 1.7, 2.8, .......,$to $100$ terms.
4. $\frac {1}{15}, \frac {1}{12}, \frac {1}{10}, .......,$ to $11$ terms.

Answer 1

(i) $2, 7, 12 ,…,$to $10$ terms
For this A.P., $a = 2$, $d = a_2 − a_1 = 7 − 2 = 5$ and $n = 10$
We know that,
$S_n = \frac n2[2a+(n-1)d]$

$S_{10} = \frac {10}{2}[2\times 2+(10-1)5]$
$S_{10} = 5[4+9\times5]$
$S_{10} = 5\times 49$
$S_{10} = 245$

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(ii) $−37, −33, −29 ,…,$ to $12$ terms
For this A.P., $a = −37$, $d = a_2 − a_1 = (−33) − (−37) = − 33 + 37 = 4, n = 12$
We know that,
$S_n = \frac n2[2a+(n-1)d]$

$S_{12} = \frac {12}{2}[2(-37)+(12-1)4]$

$S_{12} = 6[-74+11\times4]$
$S_{12} = 6[-74+44]$
$S_{12} = 6\times(-30)$
$S_{12} = -180$

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(iii) $0.6, 1.7, 2.8 ,…,$ to $100$ terms
For this A.P., $a = 0.6, d = a_2 − a_1 = 1.7 − 0.6 = 1.1$ and $n = 100$
We know that,
$S_n = \frac n2[2a+(n-1)d]$

$S_{100}$ $= \frac {100}{2}[2(0.6)+(100-1)1.1]$
S_{100}$$= 50[1.2+99\times1.1]
S_{100}$$= 50[1.2+108.9]
S_{100}$$= 50[1.2+99\times1.1]
S_{100}$$= 50[110.1]
$S_{100}$ $= 5505$

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(iv) $\frac {1}{15} , \frac {1}{12} , \frac {1}{10} ,………,$ to 11 terms

For this A.P.,
$a = \frac {1}{15}$
$n = 11$
$d = a_2-a_1$

$d = \frac {1}{12}-\frac {1}{15}$

$d = \frac {5-4}{60}$

$d = \frac {1}{60}$

We know that,

$S_n = \frac n2[2a+(n-1)d]$

$S_{11} = \frac {11}{2}[2(\frac {1}{15})+(11-1)\frac {1}{60}]$

$S_{11}$ $= \frac {11}{2}[\frac {2}{15}+\frac {10}{60}]$

$S_{11}$ $= \frac {11}{2}[\frac {2}{15}+\frac 16]$

$S_{11}$ $= \frac {11}{2}[\frac {4+5}{30}]$

$S_{11}$ $= \frac {11}{2} \times \frac {9}{30}$

$S_{11}$ $= \frac {33}{20}$