Question

How many terms of the A.P. $1,4,7$, ............ are needed to make the sum $51$?

Answer 1

We know the sum of $n$ terms of an A.P with first term $a$ and the common difference $d$ is:

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

Here, the given A.P is $1,4,7,......$ in which the first term is $a=1$, common difference is $d=4-1=3$ and the given sum is $S_n=51$, therefore,

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]\Rightarrow 51=\frac{n}{2}\left[\right(2\times 1)+(n-1\left)\right(3\left)\right]\Rightarrow 51=\frac{n}{2}(2+3n-3)\Rightarrow 51=\frac{n(3n-1)}{2}\Rightarrow 51\times 2=3n^2-n\Rightarrow 3n^2-n=102\Rightarrow 3n^2-n-102=0\Rightarrow 3n^2-18n+17n-102=0\Rightarrow 3n(n-6)+17(n-6)=0\Rightarrow 3n+17=0,n-6=0\Rightarrow n=-\frac{17}{3},n=6$

Hence, ignoring the negative values, the number of terms are $6$.