Question

The sum of the first 20 term of an AP is equal to the sum of first 30 terms show that the sum of the first 50 term of an AP is zero

Answer 1

Answer:-

Given:

Sum of first 20 terms of an AP = Sum of first 30 terms.

⟹ S₃₀ = S₂₀

We know that,

Sum of first n terms of an AP (Sₙ) = n/2 * [ 2a + (n - 1)d

So,

★ S₃₀ = 30/2 * [ 2a + (30 - 1)d ]

⟹ S₃₀ = 15 [ 2a + 29d ]

⟹ S₃₀ = 30a + 435d

★ S₂₀ = 20/2 * [ 2a + (20 - 1)d ]

⟹ S₂₀ = 10 (2a + 19d)

⟹ S₂₀ = 20a + 190d

⟹ 30a + 435d = 20a + 190d

⟹ 30a - 20a = 190d - 435d

⟹ 10a = - 245d

⟹ a = - 245d/10 -- equation (1)

Now,

We have to prove:

S₅₀ = 0

★ S₅₀ = 50/2 * [ 2a + (50 - 1)d ]

substitute the value of a from equation (1).

⟹ 0 = 25 [ 2 (- 245d/10) + 49d ]

⟹ 0 = 25 [ - 49d + 49d ]

⟹ 0 = 25(0)

⟹ 0 = 0

Hence, Proved.

Answer 2

To Prove:

• $\sf{S_{50} = 0}$

Proof:

Given that,

• Sum of first 20 term = Sum of first 30 term.

$\sf \therefore \: S_{20} = S_{30}$

$\leadsto \sf \: \frac{n}{2} \times [2a(n - 1)d ]= \frac{n}{2} \times [2a(n - 1)d ]$

[ Putting values ]

$\leadsto \sf \: \frac{20}{2} \times[ 2a + (20 - 1)d] = \frac{30}{2} \times [2a + (30 - 1)d ]\\ \\ \leadsto \sf 10 \times (2a + 19d) = 15 \times (2a + 29d) \\ \\ \leadsto \sf \: 20a + 190d = 30a + 435d \\ \\ \leadsto \sf \: 30a - 20a = 190 - 435 \\ \\ \leadsto \sf \: 10a = - 245d \\ \\ \leadsto \sf \: a = \frac{ - 245d}{10} \: \: \red \bigstar$

Now, we have to prove ;

• $\sf{S_{50} = 0}$

$\leadsto \sf \frac{n}{2} \times [2a + (n - 1)d] = 0$

$\leadsto \sf \frac{n}{2} \times [2a + (n - 1)d] = 0 \\ \\ [\: °.° \: \: \sf \:\red{a = \frac{-245d}{10}}\: \:] \\ \\ \leadsto \sf \frac{50}{2} \times [2( \frac{ - 245d}{10} ) +(50 - 1)d] = 0 \\ \\ \leadsto \sf \: 25 ( - 49d + 49d) = 0 \\ \\ \leadsto \sf 25(0) = 0 \\ \\ \leadsto \sf 0 = 0 \: \: \green \bigstar$

Hence Proved!!