Q. 2 A5.0( 1 Vote )

# Find the sums indicated below:

3 + 6 + 9 + ... + 300

Answer :

Formula used.

S_{n} = [2a + (n–1)d]

d = a_{n + 1}–a_{n}

In the sum above A.P follows

Is 3, 6, 9, ……, 300

In the following A.P

⇒ a = 3,

⇒ d = 6–3 = 3

As the last term is 300

a_{n} = a + (n–1)d

300 = 3 + (n–1)3

300 = 3 + 3n–3

3n = 300

⇒ n = = 100

S_{n} = [2a + (n–1)d]

= [2×3 + (100–1)3]

= 50[6 + 99×3]

= 50[6 + 297]

= 15150

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