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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Updated on: 10-Oct-2022

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