Question

find the sum of the following series 3+6+9+...+96​

Answer 1

it is an ap

a = 3, d = 3, an = 96

an = a+(n-1)d = 96

3+(n-1)3 = 96

(n-1)3 = 93

n-1 = 31

n = 32

put these values in

Sn= n/2[a+an]

S32 = 32/2[3+96]

S32 = 16×99 = 1584

Answer 2

Question :

Find the sum of the following series 3+6+9+...+96.

Answer :

The sum of the given series is 1584.

Given :

The series 3+6+9+...+96

To find :

The sum of the given series

Solution :

The given series is an A.P(Arithmetic Progression) series.

We know that,

an = a + (n-1)d

where an is the last term

a is the first term

n is the number of term

d is the common difference between consecutive terms

Here, d is 6-3 = 3

Hence ,

96 = 3 + (n-1) 3

=> 96 = 3 + 3n -3

=> 96 = 3n

=> n = 96/3

= 32

No.of terms is 32

We know,

sum of A.P series

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

= 32/2 [ 2 × 3 + ( 32 - 1) 3]

= 16 [ 6 + 31×3]

= 16 [ 6 + 93]

= 16 × 99

= 1584

Hence, the sum of the given series is 1584.

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