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 termsHere, $$a=63, d=60-63=-3$$ 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.MathematicsRS AgarwalStandard X

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