Find the sum of the following AP’s.
(i) 2, 7, 12… to 10 terms
(ii) –37, –33, –29… to 12 terms
(iii) 0.6, 1.7, 2.8… to 100 terms
(iv) \({{1} \over {15}},{{1} \over {12}},{{1} \over {10}}…\)to 11 terms

The formula for finding the sum of A.P. is \(S_n = {{n} \over {2}}[2a + (n - 1)d]\)

Where a= first term, d = common difference and n = number of terms

(i)2, 7, 12… to 10 terms
Here a = 2,
d = \(7 - 2 = 5\)
and n = 10
\(S_n = {{10} \over {2}}(4 + (10 - 1)5) \)
\(= 5(4 + 45) = 5(49) = 245\)


(ii) –37, –33, –29… to 12 terms

Here a = -37, d = \(-33 - (-37) = 4\) and n = 12

\(S_n = {{12} \over {2}}(-74 + (12 - 1)4) \)
\( = 6(-74 + 44) = 6(-30) = -180\)


(iii) 0.6, 1.7, 2.8… to 100 terms

Here a = 0.6, d = \(1.7 - 0.6 = 1.1\) and n = 100
\(S_n = {{100} \over {2}}(1.2 + (100 - 1)1.1) \)
\( = 50(1.2 + 108.9) \)
\( = 50(110.1) = 5505\)


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

Here a = \({{1} \over {15}}\), d = \({{1} \over {12}} - {{1} \over {15}} = {{1} \over {60}}\) and n = 100

\(S_n = {{11} \over {2}}({{2} \over {15}} + (11 - 1){{1} \over {60}}) \)
\( = {{11} \over {2}}({{2} \over {15}} + {{1} \over {6}}) = {{11} \over {2}}({{9} \over {30}}) = {{33} \over {20}}\)