Question

If 10 times of 10th term of an AP is equal to 15 times the 15 th term ,

show that its 25 th is equal to zero.
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Answer 1

Answer:

Step-by-step explanation:

Given:

• 10 times the 10th term of an A.P = 15 times 15th term of the A.P

To Prove:

• The 25th term of the A.P = 0

Proof:

Let the first term of the A.P be a₁.

Let the common difference of the A.P be d.

The nth term of an A.P is given by,

aₙ = a₁ + (n - 1) × d

where aₙ is the nth term

a₁ is the first term

d is the common difference

Hence,

10th term of the A.P (a₁₀) = a₁ + 9 d------(1)

15th term of the A.P (a₁₅) = a₁ + 14 d------(2)

But by given,

10 × a₁₀ = 15 × a₁₅

Substitute equation 1 and 2,

10 × (a₁ + 9d) = 15 × (a₁ + 14d)

10a₁ + 90d = 15a₁ + 210d

10a₁ - 15a₁ = 210d - 90d

-5a₁ = 120d

a₁ = -24 d----(3)

Now the 25th term of an A.P is given by,

a₂₅ = a₁ + 24d

Substitute the value of a₁ from equation 3,

a₂₅ = -24d + 24d

a₂₅ = 0

Therefore the 25th term of the A.P is 0.

Hence proved.

Answer 2

Question

if 10 times of 10th term of an AP is equal to 15 times the 15th term.

Show that it's 25th term is equal to zero.

Given

• 10 times of 10th term of an A.P is equal to 15th times the 15th term.

Required to find

To show that it's 25th term is equal to zero

Solution

We know that the nth term of the arithmetic progression is given by a + ( n - 1 )d

Given that the 10 times the 10th term is equal to 15th times the 15th term.

Therefore, 10 ( 10th term ) = 15 ( 15th term )

$\sf \implies 10(a + (10 - 1)d) = 15(a + (15 - 1)d)$

$\\ \sf \implies \: 10(a + 9d) = 15(a + 14d)$

$\\ \implies \sf \: 10a + 90d = 15a + 210d$

$\\ \implies \sf \: 15a - 10a = 90d - 210d$

$\\ \implies \sf \: 5a = - 120d$

$\sf \implies \color{cyan} \: a = - 24d......(1)$

The 25th term is a + (25 - 1 )d = a + 24d = -24d + 24d = 0

$\tt \color{purple} \: therefore \: the \: {25}^{th} term \: \: of \: the \: A.P \: is \: zero$