Find the sum : 34 + 32 + 30 + .... + 10
286
34+32+30+....+10
This sequence is an AP.
Here, a=34,d=32−34=−2.
Let the number of terms of the AP be n.
We know that
an=a+(n−1)d.
⇒10=34+(n−1)(−2)
⇒(n−1)(−2)=−24
⇒n−1=242=12
⇒n=13
Also, we know that the sum to n terms of an AP with first term a and last term l is given by
Sn=n2(a+l).
∴S13=132(34+10)
S13=286
Hence, the required sum is 286.