the sum of the first n terms of an AP is 3n2+5n.write A.P hence find its 16th term
Answers
Answered by
6
hello
The sum of n terms of an A.P. = 3n2+5n
then sum of n-1 terms= 3(n-1)2+5(n-1)
So nth term of A.P will be Tn=3n2+5n- 3(n-1)2-5(n-1)= 3n2+5n-3n2-3+6n-5n+5=6n+2
in this way by putinng values of n= 1,2,3,4,5,6,7,_______n-1, n
we will get the AP as 8,14,20,26,32,38,44,____and the 16th term will be= 6*16+2=98
The sum of n terms of an A.P. = 3n2+5n
then sum of n-1 terms= 3(n-1)2+5(n-1)
So nth term of A.P will be Tn=3n2+5n- 3(n-1)2-5(n-1)= 3n2+5n-3n2-3+6n-5n+5=6n+2
in this way by putinng values of n= 1,2,3,4,5,6,7,_______n-1, n
we will get the AP as 8,14,20,26,32,38,44,____and the 16th term will be= 6*16+2=98
Answered by
7
S(n)=5n^2 - 3n
S(1) = t1 = 2
S(2) = t1+t2 = 14 → t2 = 12 → d= 10
t(16) = 2+9(10) = 92
S(1) = t1 = 2
S(2) = t1+t2 = 14 → t2 = 12 → d= 10
t(16) = 2+9(10) = 92
Similar questions