CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

How many terms of the AP $$63, 60, 57, 54,...$$ must be taken so that their sum is $$693$$? Explain the double answer.


Solution

AP is $$63, 60, 57, 54,....$$
Consider
$$a =$$ First term
$$d =$$ Common difference
$$n =$$ number of terms
Here, $$a=63, d=60-63=-3$$ and sum $$=S_n=693$$

$$S_n=\dfrac{n}{2}[2a+(n-1)d]\\$$
$$693=\dfrac{n}{2}[2\times 63+(n-1)(-3)]$$
$$693\times 2=n(126-3n+3)$$
$$1386=n(129-3n)$$
$$1386=129n-3n^2$$
$$3n^2-129n+1386=0$$
$$n^2-43n+462=0$$
Which is a quadratic equation
$$n^2-21n-22n+462=0$$
$$n(n-21)-22(n-21)=0$$
$$(n-21)(n-22)=0$$
Either, $$n-21=0$$, then $$n=21$$
or $$n-22=0$$, then $$n=22$$

Number of terms$$=21$$ or $$22$$
$$T_{22}=a+(n-1)d$$
$$=63+(22-1)(-3)$$
$$=63+21\times (-3)=63-63=0$$
Which shows that, $$22$$th term of AP is zero.
Number of terms are $$21$$ or $$22$$. So there will be no effect on the sum.

Mathematics
RS Agarwal
Standard X

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
Same exercise questions
View More


similar_icon
People also searched for
View More



footer-image