How many terms of the A.P. 63, 60, 57, . . . must be taken so that the

1.	How many terms of the A.P. 63, 60, 57, . . . must be taken so that their sum is 693?
Answer» Given A.P. is 63, 60, 57,… We know that, S_n = \(\frac{n}{2}\)[2a + (n − 1)d] Here we have, the first term (a) = 63 The sum of n terms (S_n) = 693 Common difference of the A.P. (d) = a₂ – a₁ = 60 – 63 = –3 On substituting the values in S_nwe get ⟹ 693 = \(\frac{n}{2}\)[2(63) + (n − 1)(−3)] ⟹ 693 = \(\frac{n}{2}\)[163+(−3n + 3)] ⟹ 693 = \(\frac{n}{2}\)[129 − 3n] ⟹ 693(2) = 129n – 3n² Now, we get the following quadratic equation. ⟹ 3n² – 129n + 1386 = 0 ⟹ n² – 43n + 462 Solving by factorisation method, we have ⟹ n² – 22n – 21n + 462 = 0 ⟹ n(n – 22) -21(n – 22) = 0 ⟹ (n – 22) (n – 21) = 0 Either, n – 22 = 0 ⟹ n = 22 Or, n – 21 = 0 ⟹ n = 21 Now, the 22^nd term will be a₂₂ = a₁ + 21d = 63 + 21( -3 ) = 63 – 63 = 0 So, the sum of 22 as well as 21 terms is 693. Therefore, the number of terms (n) is 21 or 22.

How many terms of the A.P. 63, 60, 57, . . . must be taken so that their sum is 693?

Answer»

Given A.P. is 63, 60, 57,…

We know that,

S_n = \(\frac{n}{2}\)[2a + (n − 1)d]

Here we have,

the first term (a) = 63

The sum of n terms (S_n) = 693

Common difference of the A.P. (d) = a₂ – a₁ = 60 – 63 = –3

On substituting the values in S_nwe get

⟹ 693 = \(\frac{n}{2}\)[2(63) + (n − 1)(−3)]

⟹ 693 = \(\frac{n}{2}\)[163+(−3n + 3)]

⟹ 693 = \(\frac{n}{2}\)[129 − 3n]

⟹ 693(2) = 129n – 3n²

Now, we get the following quadratic equation.

⟹ 3n² – 129n + 1386 = 0

⟹ n² – 43n + 462

Solving by factorisation method, we have

⟹ n² – 22n – 21n + 462 = 0

⟹ n(n – 22) -21(n – 22) = 0

⟹ (n – 22) (n – 21) = 0

Either, n – 22 = 0 ⟹ n = 22

Or, n – 21 = 0 ⟹ n = 21

Now, the 22^nd term will be a₂₂ = a₁ + 21d = 63 + 21( -3 ) = 63 – 63 = 0