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

(ii)Explain the double answer

Question Number 20 Of Exercise 11 C of RS Aggarwal Solution.

Deepak BoraNewbie

# 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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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.

Verification of the Answer

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