Let’s take the first term as a and the common difference to be d
Given that,
a5 = 30 and a12 = 65
And, we know that an = a + (n – 1)d
So,
a5 = a + (5 – 1)d
30 = a + 4d
a = 30 – 4d …. (i)
Similarly, a12 = a + (12 – 1) d
65 = a + 11d
a = 65 – 11d …. (ii)
Subtracting (i) from (ii), we have
a – a = (65 – 11d) – (30 – 4d)
0 = 65 – 11d – 30 + 4d
0 = 35 – 7d
7d = 35
d = 5
Putting d in (i), we get
a = 30 – 4(5)
a = 30 – 20
a = 10
Thus for the A.P; d = 5 and a = 10
Next, to find the sum of first 20 terms of this A.P., we use the following formula for the sum of n terms of an A.P.,
Sn =\(\frac{ n}{2}\)[2a + (n − 1)d]
Where;
a = first term of the given A.P.
d = common difference of the given A.P.
n = number of terms
Here n = 20, so we have
S20 = \(\frac{20}{2}\)[2(10) + (20 − 1)(5)]
= (10)[20 + (19)(5)]
= (10)[20 + 95]
= (10)[115]
= 1150
Hence, the sum of first 20 terms for the given A.P. is 1150.