The third term of an A.P. is 7 and the seventh term exceeds three times the third term by 2.

Solution:

In the given problem, let us take the first term as a and the common difference as d.

Here, we are given that,

$a_{3}=7$...........(1)

$a_{7}=3 a_{3}+21$.........(2)

So, using (1) in (2), we get,

$a_{7}=3(7)+2$

$=21+2$

$=23$...........(3)

Also, we know,

$a_{n}=a+(n-1) d$

For the $3^{\text {th }}$ term $(n=3)$,

$a_{3}=a+(3-1) d$

$7=a+2 d$(Using 1)

 

$a=7-2 d$......(4)

Similarly, for the $7^{\text {th }}$ term $(n=7)$,

$a_{7}=a+(7-1) d$

$24=a+6 d$ (Using 3)

$a=24-6 d$...............(5)

Subtracting (4) from (5), we get,

$a-a=(23-6 d)-(7-2 d)$

$0=23-6 d-7+2 d$

 

$0=16-4 d$

$4 d=16$

$d=4$

Now, to find a, we substitute the value of d in (4),

$a=7-2(4)$

$a=7-8$

 

$a=-1$

So, for the given A.P $d=4$ and $a=-1$

So, 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.,

$S_{\pi}=\frac{n}{2}[2 a+(n-1) d]$

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, using the formula for n = 20, we get,

$S_{20}=\frac{20}{2}[2(-1)+(20-1)(4)]$

$=(10)[-2+(19)(4)]$

$=(10)[-2+76]$

$=(10)[74]$

$=740$

Therefore, the sum of first 20 terms for the given A.P. is $S_{20}=740$.