Question

[Class 10, Arithmetic Progression]
Q. How many terms of the AP 63, 60, 57, 54... must be taken so that their sum is 636?​

Answer 1

Step-by-step explanation:

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

Sп = 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