Q : 1      Find the sum of the following APs:

              (iv)   \small \frac{1}{15},\frac{1}{12},\frac{1}{10},...,  to   \small 11  terms.

Answers (1)

Given AP is 
 \small \frac{1}{15},\frac{1}{12},\frac{1}{10},...,  to   \small 11  terms.
Here, a = \frac{1}{15} \ and \ n = 11
And
d = a_2-a_1=\frac{1}{12}-\frac{1}{15}= \frac{5-4}{60}= \frac{1}{60}
Now, we know that 
S = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S = \frac{11}{2}\left \{ 2\times \frac{1}{15} +(11-1)(\frac{1}{60})\right \}
\Rightarrow S = \frac{11}{2}\left \{ \frac{2}{15} +\frac{1}{6}\right \}
\Rightarrow S = \frac{11}{2}\left \{ \frac{9}{30}\right \}
\Rightarrow S =\frac{99}{60}= \frac{33}{20}
Therefore, the sum of AP  \small \frac{1}{15},\frac{1}{12},\frac{1}{10},...,  to   \small 11  terms. is \frac{33}{20}

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