Arithmetical Progression

Today i am solve some problem related to A.P.
1. find the 11th of a A.P. which first term is 3 and common difference is 5.
Ans: Give that first term a=3 and common difference is 5
so nth term = a+(n-1)d ;
putting the value of a, d and n in above equation , we get;
a=3; d=5 and n=11
11th term = 3+ (11-1)5
= 3+ 10*5
= 3+50
= 53 ans
Q 2. A arithmetical progression is as follows 10, 6,2,...............
Find the 10th term and the addition of first ten terms.
Ans: Here given series is 10,6,2,.........
so first term a=10 and common difference = 6- 10= -4 ( astounding to definition )
so the 10th term = 10+ (10-1)(-4)
= 10 + 9(-4)
= 10-36 [we know (+)*(-)= -]
= -26
Now the addition
first term a= 10, d= -4 and number of term n =10
First method,
summation of first n term s= n/2[2a+ (n-1)d]
Putting the value of a, d, and n, in above equation , we get
s= 10/2[2*10 +(10-1)*(-4)];
= 5[20 -36];
= 5(-16)= -80
so the summation of ten term is -80 .
Second method
sum of n term =n/2[first term + last term];
s= 10/2[ 10 +(-26)] [ here last term is -26]
= 5[10 - 26];
= 5( -16) = -80 .

NOTE: in this type of problem any three variable are give out of four variable such as a ,d ,n and s . and uyou have to find the other variable . put the all value in given equation of summation and find the unknow .
Thanks