Question

is 33 a term of the series 3+6+9...+12 ? find the sum of series

Answer 1

A.P. = 3, 6, 9, 12, ....

First term, a = 3

Second term, $\sf{a_2}$ = 6

Common Difference, d = $\sf{a_2}$ - a = 6 - 3 = 3

$\sf{a_n}$ = 33

${\boxed{\sf{a_n = a + (n - 1) d}}}$

33 = 3 + ( n - 1 ) ( 3 )

33 - 3 = ( n - 1 ) ( 3 )

30 = 3n - 3

3n = 30 + 3

3n = 33

n = 33 / 3

n = 11

Hence, we conclude that $<b>33 is a term of the given A.P.</b>$

Now,

A.P. = 3 + 6 + 9 + 12

n = 4

${\boxed{\sf{S_n = {\dfrac{n}{2}} (a + a_n)}}}$

$\sf{S_4}$ = ( 4 / 2 ) ( 3 + 12 )

= 2 ( 15 )

= 30

Hence, $<b>Sum of series is 30.<b>$