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# 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

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