Question

10. Find the number of terms of A.P. $54,51,48,......$ so that their sum is $513$

Answer 1

Step 1: Note the given data

Given Arithmetic progression is $54,51,48,......$

First-term $t_1=a=54$

Second-term $t_2=51$

Third-term $t_3=48$

Step 2: Finding the common difference of A.P.

The formula for finding the common difference in A.P. $={(n+1)}^{th}term-n^{th}term$

Common difference $d=51-54$

$=-3$

Similarly, $d=51-54=48-51=-3$

Step 3: Finding the $n^{th}$ term of A.P.

Let the last be the $n^{th}$ term

The general formula for finding the $n^{th}$ term of A.P. $=a+(n-1)d$

$n^{th}$ term $=54+(n-1)\times (-3)$

$$\begin{gathered} =54-3n+3 \\ =57-3n \end{gathered}$$

Step 4: Finding the sum of A.P.

The general formula for finding the sum of A.P. $=\frac{n}{2}(t_1+t_n)$

The sum of the A.P. $=\frac{n}{2}(54+57-3n)$

$=\frac{n}{2}(111-3n)$

Step 5: Equating both the sums and we get

$\frac{n}{2}(111-3n)=513$

$$\begin{gathered} \Rightarrow n(111-3n)=513\times 2 \\ \Rightarrow 111n-3n^2=1026 \end{gathered}$$

$\Rightarrow 3n^2-111n+1026=0$

$\Rightarrow 3(n^2-37n+342)=0$

Divide both sides of the above equation by $3$

$\Rightarrow n^2-37n+342=0$

$\Rightarrow n^2-(18+19)n+342=0$

$\Rightarrow n^2-18n-19n+342=0$

$\Rightarrow n(n-18)-19(n-18)=0$

$\Rightarrow (n-18)(n-19)=0$

Either $n-18=0$ or, $n-19=0$

Either $n=18$ or, $n=19$

Hence, the number of terms of A.P. is $18$ or $19$