# Calculate the sum of 1001 terms of the series 3,5,7,...

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

3, 5, 7, ...., a1001  is an A.P  such that:

a1= 3

a2= 3+ 1*2

a3= 3+ 2*2

a4= 3+  3*2

.....

a1001 = 3+ 1000*2 = 2003

r= 2

==> S1001 = 3 + 5 + 7 + ...+ a1001

We know that:

Sn = (a1+an)*n/2

Sn = (3+ 2003)*1001/2

= (2006)(1001)/2= 1004003.

==: Sn = 1004003

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

The series we have is an AP : 3, 5, 7 ...

We see that the first term is 3 and the common difference is 7-5 = 5-3 = 2.

Now we need to find the sum of 1001 terms.

For an AP the relation for the sum of n terms terms is [2a + (n-1)*d] n/2

a= 3 and d = 2

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

=> [2*3 + (1001-1)*2]* 1001/2

=> [6 + (1000)*2] *1001/2

=> [6 + 2000]*1001 /2

=> 2006*1001/2

=> 1003*1001

Therefore the required sum is 1004003

neela | High School Teacher | (Level 3) Valedictorian

Posted on

The terms of the series are 3,5,7,.....

Therefore the common difference d of the series  d = 7-5=5-3 = 2.

Thefirst term of the series a1 = 3

Therefore the n the term of the series a1 +(n-1)d = 3 +(n-1)*2.

Therefore the 2001 th term = a1 +(n-1)d = 3+(2001-1)*2 = 4003.

Therefore the sum of the series Sn = (a1 +a1+(n-1)d }n/2 = {First term +Last term }number of terms/2

S2001 = (3+4003)(2001)/2 = 4006*2001/2 = 4008003

Therefore the sum of 2001 terms of the series 3,5,7,9,.....4003 is 4008003.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We notice that if we'll calculate the difference between 2 consecutive terms of the given series, we'll obtain the same value each time:

5 - 3 = 7 - 5 = ...... = 2

So, the given series is an arithmetic progression whose common difference is d = 2.

We can calculate the sum of n terms of an arithmetic progression in this way;

Sn = (a1 + an)*n/2

a1 - the first term of the progression

a1 = 3

an - the n-th term of the progression

an = a1001

a1001 = a1 + (1001-1)*d

a1001 = 3 + 1000*2

a1001 = 3 + 2000

a1001 = 2003

n = 1001 - the number of terms

S1001 = [a1 + a1 + (1001-1)*d]*1001/2

S1001 = (3 + 2003)*1001/2

S1001 = 2006*1001/2

S1001 = 1003*1001

The sum of the 1001 terms of the series 3,5,7,... is:

S1001 = 1004003