Question

the sum of third term and eleventh term of an A.P. is 34. find the sum of its 13 terms.

Answer 1

we know that
third term = a+2d-------a
and,
11th term = a+ 10d--------b
now ,
a+b=2a+12d=34----------- [given]
now sum of 13 terms is
Sn=S13 = 13/2×[2a+12d]
S13=13/2×34 ---------[2a+12d=34]

therefore S13= 13 ×17= 221
answer=221

Answer 2

Answer:

The sum of 13 terms of the AP is 221.

Step-by-step explanation:

The nth term of an AP is defined as

$a_n=a+(n-1)d$

where, a is first term and d is common difference.

Third term of AP is

$a_3=a+(3-1)d=a+2d$

Eleventh term of AP is

$a_{11}=a+(11-1)d=a+10d$

It is given that the sum of third term and eleventh term of an A.P. is 34.

$a_3+a_{11}=34$

$a+2d+a+10d=34$

$2a+12d=34$                    ..... (1)

The sum of n terms of an AP is

$S_n=\frac{n}{2}[2a+(n-1)d]$

The sum of 13 terms of the AP is

$S_{13}=\frac{13}{2}[2a+(13-1)d]$

$S_{13}=\frac{13}{2}[2a+12d]$

Using (1), we get

$S_{13}=\frac{13}{2}\times 34=221$

Therefore the sum of 13 terms of the AP is 221.