Question

sum of 9+11+13+............+29​

Answer 1

Answer :

209

Note :

★ A.P. (Arithmetic Progression) : A sequence in which the difference between the consecutive terms are equal is said to be in A.P.

★ If a1 , a2 , a3 , . . . , an are in AP , then

a2 - a1 = a3 - a2 = a4 - a3 = . . .

★ The common difference of an AP is given by ; d = a(n) - a(n-1) .

★ The nth term of an AP is given by ;

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

★ If a , b , c are in AP , then 2b = a + c .

★ The sum of nth terms of an AP is given by ; S(n) = (n/2)×[ 2a + (n - 1)d ] .

or S(n) = (n/2)×(a + l) , l is the last term .

★ The nth term of an AP can be also given by ; a(n) = S(n) - S(n-1) .

Solution :

Here ,

The given series is ; 9 + 11 + 13 + . . . + 29

Clearly ,

The given series is an arithmetic series .

Here ,

• First term , a = 9

• Common difference , d = 11 - 9 = 2

• Least term , l = 29

Now ,

=> l = 29

=> a(n) = 29

=> a + (n - 1)d = 29

=> 9 + (n - 1)•2 = 29

=> 2(n - 1) = 29 - 9

=> 2(n - 1) = 20

=> n - 1 = 20/2

=> n - 1 = 10

=> n = 10 + 1

=> n = 11

Now ,

We know that ,

S(n) = (n/2)•( a + l)

Thus ,

=> S(11) = (11/2)•(a + l)

=> S(11) = (11/2)•(9 + 29)

=> S(11) = (11/2)•(38)

=> S(11) = 11•19

=> S(11) = 209

Hence ,

9 + 11 + 13 + . . . + 209 .