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# Find the number of terms of the AP 63,60,57,……so that their sum is 693. Explain the double answer.

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This is the Important question based on Arithmetic progression Chapter of R.S Aggarwal book for ICSE & CBSE Board.
Here you have to find the
(i) Number of terms so that their sum is Given
Question Number 20 Of Exercise 11 C of RS Aggarwal Solution.

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1. Given, AP 63, 60, 57
where, a = 63
and the difference (d) = 60 – 63 = -3
also given that Sп = 693
∴ to find a,
we know
S_n = n/2 [ 2a + (n -1) d ]
By substituting the values of a, d and Sп we get;
693 = n/2 [ 2 ×63 + (n – 1) – 3 ]
693 = n/2 [ 126 – 3n + 3 ]
693 = n/2 [ 129 – 3n ]
693 = 129n/2 – 3n²/2
693 × 2 = 129n – 3n²
1386 = 129n -3n²
1386 – 129n + 3n² = 0
By dividing the whole equation by 3
we get,
1386/3 – 129n/3 + 3n²/3 = 0/3
462 – 43n + n² = 0
ie; n² – 43n + 462 = 0
using factorisation method :–
sum = – 43 and product = 462
∴ the numbers are -21 and -22
So by splitting the middle term we get;
( n² – 21n ) ( – 22n + 462 ) = 0
n ( n – 21 ) – 22 ( n – 21 ) = 0
( n – 21 ) ( n – 22 ) = 0
∴ n = 21 and n = 22
ie; We get the sum of the given AP as 693 when we take first 21 terms of it or 22 terms of the same AP.
First take n as 21, the S₂₁ = 21/2 ( a + a₂₁ )
a₂₁ = a + ( 21 – 1 ) d
= 63 + [ 20 × ( – 3 ) ]
= 63 – 60
a₂₁ = 3
∴ s₂₁ = 21/2 [ 63 + 3 ]
= 21/2 × 66
s₂₁ = 693
So the condition is satisfied for when n = 21
Now check for when n = 22
s₂₂ = 22/2 ( a + a₂₂ )
a₂₂ = a + ( 22 – 1 ) d
= 63 + [ 21 × ( -3 ) ]
= 63 – 63
a₂₂ = 0
We know
s₂₂ = s₂₁ + a₂₂
= 693 + 0
= 693
∴ the condition is satisfied in both the cases
so n = 21 or n = 22

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