The third term of an A.P. is 7 and the seventh term exceeds three times the third term by 2. Find the first term, the common difference and the sum of first 20 terms.
Given:
The third term of an A.P. is 7 and the seventh term exceeds three times the third term by 2.
To do:
We have to find the first term, the common difference and the sum of first 20 terms.
Solution:
Let the first term, common difference and the number of terms of the given A.P. be $a, d$ and $n$ respectively.
We know that,
nth term of an A.P. $a_n=a+(n-1)d$
Therefore,
First term $a_1=a$
$a_{3}=a+(3-1)d$
$7=a+2d$.....(i)
$a_{7}=a+(7-1)d$
$=a+6d$....(ii)
According to the question,
$a_{7}=3a_3+2$
$a+6d=3(a+2d)+2$
$a+6d=3a+6d+2$
$3a-a=-2$
$2a=-2$
$a=-1$....(iii)
Substituting $a=-1$ in (i), we get,
$7=-1+2d$
$2d=7+1$
$d=\frac{8}{2}$
$d=4$
The sum of $n$ terms of an A.P. $S_n=\frac{n}{2}[2a+(n-1)d]$
$S_{20}=\frac{20}{2}[2(-1)+(20-1)4]$
$=10(-2+76)$
$=10(74)$
$=740$
The first term is $-1$, the common difference is $4$ and the sum of the first 20 terms is $740$.
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