Question

29. How many terms of the AP: 20, 18.16, ......
must be taken so that their sum is 110?​

Answer 1

A.P → 20, 18, 16, ...

Here,

• First term = a = 20
• Common Difference = d = a₂ - a₁ = 18 - 20 = -2
• Sum of n terms = Sₙ = 110

We need to find the number of terms that is n

$\sf{S_n = \dfrac{n}{2} \times (2a + (n - 1)d)}\\\\\\\implies \sf{ 110 = \dfrac{n}{2} \times (2 \times 20 + (n - 1) \times -2)}\\\\\\\implies \sf{110= \dfrac{n}{2} \times (40 + (n - 1) \times -2)}\\\\\\\implies \sf{110 = \dfrac{n}{2} \times (40 - 2n + 2)}\\\\\\\implies \sf{110 = \dfrac{n}{2} \times (42-2n)}\\\\\\\implies \sf{110 = \dfrac{n}{2} \times 2(21 - n)}\\\\\\\implies \sf{110 = n(21 - n)}\\\\\\\implies \sf{110 = 21n - {n}^{2}}\\\\\\\implies \sf{n^2 + 110 - 21n = 0}$

⇒ n² - 21n + 110 = 0

⇒ n² - 10n - 11n + 110 = 0

⇒ n(n - 10) - 11(n - 10) = 0

⇒ (n - 11)(n - 10) = 0

⇒ n = 10 or 11

∴ 10 or 11 terms must be taken