Question

in certain AP the 24th term is twice the 10th term. Prove that the 72nd term is twice the 34th term.​

Answer 1

Step-by-step explanation:

Given: a₂₄=2a₁₀

Prove that: a₇₂=2a₃₄

Lets's find the given terms of AP:

• a₂₄= a₁+23d
• a₁₀= a₁+9d

Comparing the 2 equations above:

• a₁+23d= 2(a₁+9d)
• a₁=5d

Now, next 2 terms:

• a₇₂= a₁+71d = 5d+71d= 76d

• a₃₄= a₁+33d= 5d+33d= 38d

As per question, 72nd term is twice the 34th term, let's check:

• a₇₂= 76d
• a₃₄= 38d
• a₇₂/a₃₄= 76d/38d = 2

So, proved that  72nd term is twice the 34th term

Hope it helps

Answer 2

Given:-

• The 24th term is twice the 10th term.

Prove that:-

• The 72nd term is twice the 34th term.

Solutions:-

The 24th term is twice the 10th term.

• Let the first term of the Ap as a and the common difference as d.

We know,

• an = a + (n - 1)d

For 10th term (n = 10)

• a10 => a + (n - 1)d

=> a + 9d

For 24th term (n = 24)

• a24 => a + (24 - 1)d

=> a + 23d

Now, we are given that a24 = 2a10

So, we get,

=> a + 23d = 2(a + 9d)

=> a + 23d = 2a + 18d

=> 23d - 18d = 2a - a

=> 5a = a ...........(1).

The 72nd term is twice of 34th term.

For 34th term (n = 34)

• a34 => a + (34 - 1)d

=> a + 33d. (using Eq. 1)

=> 5d + 33d

=> 38d

For 72nd term (n = 72)

• a72 => a + (72 - 1)d

=> a + 71d. (using Eq. 1)

=> 5d + 71d

=> 76d

=> 2(38d)

Hence, The Ap a72 term is twice of the 34th.