Question

the first and last term of an ap are 4 and 81respectively if the common differnce 7 how many terms ate tere in the ap and whay is there sum

Answer 1

a=4 an=81 d=7 an=a+(n-1)d 81=4+(n-1)7 ;77=(n-1)7;11=n-1;n=12

Answer 2

a=4
an=81
D=7
an=a+(n-1)D
81=4+(n-1)7
7n-7+4=81
7n=81+3
n=84/7
n=12,thus no. of terms=12

sum of A.P.=n/2[a+an]
=12/2(4+81)
=6×85
=510