Question

Find the sum of first 50 term's od A.P. with pattern 3,6,9,12...

Answer 1

For the pattern we have,

First term (a) = 3

Common difference (d) = 3

We know that,

$S _{n} = \frac{n}{2} (2a + (n - 1)d)$

$= > S _{50} = \frac{50}{2} (6 + (49 \times 3) \\ = > S _{50} = 25 \times (6 + 147) \\ = > S _{50} = 25 \times 153 = 3825$


Hence, the sum of first 50 term's of A.P is 3,825.

Answer 2

comment difference is(d)

6-3=3

9-6=3

and first term of the AP is (a)

=3

Sn=n/2(2a+(n-1)d)

where n is 50

S50=50/2(2(3)+(50-1)3)

S50=25(6+(49)3)

S50=25(6+147)

S50=25(153)

S50=3825

hope this answers will help you