Question

Sum of first and 25th terms of an arithmetic sequence is 70.
(i) What is the sum of 4th and 22nd terms of this sequence ?

Hint:

Remembering the property ap + aq = aᵣ + aₛ if p+q=r+s makes this problem a simple observation. It saves a lot of time compared to expanding the terms using the aₙ formula.

Answer 1

We are given the sum of the 1st and 25th terms of an AP. We need to find the sum of the 4th and 22nd terms.
Given: a₁ + a₂₅ = 70.

A key property of an arithmetic progression is that if the sum of the indices of two pairs of terms is equal, then the sum of the terms in those pairs is also equal.
That is, if p + q = r + s, then ap + aq = aᵣ + aₛ.

Method 1: Using the property.
We are given the sum of the 1st and 25th terms. The sum of their indices is 1 + 25 = 26.
We need to find the sum of the 4th and 22nd terms. The sum of their indices is 4 + 22 = 26.
Since the sum of the indices is the same (1+25 = 4+22), the sum of the terms must also be the same.
Therefore, a₄ + a₂₂ = a₁ + a₂₅ = 70.

Method 2: Using the general formula.
Let a be the first term and d be the common difference.
Given: a₁ + a₂₅ = 70.
a + (a + (25-1)d) = 70 a + a + 24d = 70 2a + 24d = 70 --- (1)
Now, let's find the sum of the 4th and 22nd terms:
a₄ + a₂₂ = (a + (4-1)d) + (a + (22-1)d)
a₄ + a₂₂ = (a + 3d) + (a + 21d)
a₄ + a₂₂ = 2a + 24d
From equation (1), we know that 2a + 24d = 70.
Therefore, the sum of the 4th and 22nd terms is 70.

The sum of the 4th and 22nd terms is 70.