How many terms of the A.P. $63, 60, 57, ………$ must be taken so that their sum is 693?

Given:

Given A.P. is $63, 60, 57, ………$
To do:

We have to find the number of terms that must be taken so that their sum is 693.
Solution:

Let the number of terms be \( n \).

First term \( (a)=63 \)

Common difference \( (d)=60-63=-3 \)

We know that,

\( S_{n}=\frac{n}{2}[2 a+(n-1) d] \)
\( \Rightarrow 693=\frac{n}{2}[2 \times 63+(n-1) \times(-3)] \)
\( \Rightarrow 693=\frac{n}{2}[126-3 n+3] \)
\( \Rightarrow 1386=n(-3 n+129) \)
\( \Rightarrow 1386=-3 n^{2}+129 n \)
\( \Rightarrow 3 n^{2}-129 n+1386=0 \)

\( \Rightarrow 3(n^{2}-43 n+462)=0 \)
\( \Rightarrow n^{2}-43 n+462=0 \)
\( \Rightarrow n^{2}-21 n-22 n+462=0 \)
\( \Rightarrow n(n-21)-22(n-21)=0 \)
\( \Rightarrow(n-21)(n-22)=0 \)
This implies,

\( n-21=0 \) or \( n-22=0 \) 

\( n=21 \) or \( n=22 \)

The number of terms to be taken is 21 or 22.   

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Updated on: 10-Oct-2022

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