What is the sum of the 20th term of an ap when the fifth term is 9 and twelfth term is 30

Answer 1

5th term is 9 ------ a + 8d = 9
12 term is 30 ----- a + 11d = 30

subtract them:
3d = 21
d = 7
into a + 8d = 9 gives us a = -47

The 20th term is a + 19d = -47 + 19(7) = 86
sum(20) = 10(first + last) = 10(-47 + 86) = ...
or
= (20/2)( 2(-47) + 19(7) ) = ...
using sum(n) = (n/2)(2a + (n-1)d )

Answer 2

a5 = a1 + 4 d = 9

a12 = a1 + 11 d = 30

Now you must solve system:

a1 + 4 d = 9

a1 + 11 d = 30

The solution is:

a1 = - 2 , d = 3

Sum of the first n terms of an AP::

Sn = n ( a1 + an ) / 2

In this case:

n = 20

a20 = a1 + 19 d = - 3 + 19 * 3

a20 = - 3 + 57

a20 = 54

S20 = 20 ( a1 + a20 ) / 2

S20 = 20 ( - 3 + 54 ) / 2

S20 = 20 * 51 / 2

S20 = 510