Question

3)Find the sum of 3+6+9+....+120​

Answer 1

Solution

a=3

d=6-3=3

l=120

let 120 is the n'th term

now..

a(n)=a+(n-1)d

=>120=3+(n-1)3

=>n=40

therefore..

$s(40) = \frac{40(3 + 120)}{2} \\ = 123 \times 20 \\ = 2460$

Answer 2

The sum of the series is 2460

Step-by-step explanation:

The given series, 3 + 6 + 9 + .... + 120​

This is Arithmetic sequence

First term, a = 3

Common difference, d = 3  (difference of consecutive term)

last term, l = 120

The formula of last term.

$a_n=a+(n-1)d$

$120=3+(n-1)\cdot 3$

$3(n-1)=117$

$n-1=39$

$n=40$

Sum of sequence formula, $S_n=\dfrac{n}{2}(a+l)$

$Sum=\dfrac{40}{2}(3+120)$

$Sum=20\times 123$

$Sum=2460$

Hence, the sum of the series is 2460

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Sum of arithmetic series

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