Question

Let $a_1, a_2, a_3, \dots$ be in A.P. and $h_1, h_2, h_3, \dots$ in H.P. If $a_1 = 2 = h_1$ and $a_{30} = 25 = h_{30}$ then $(a_{24} + a_{14}h_{17})$ equal to :

• A. 50
• B. 100
• C. 200
• D. 400

Answer 1

Answer: (A)

50

Answer 2

The problem involves terms from an Arithmetic Progression (A.P.) and a Harmonic Progression (H.P.). We are given initial and final terms for both sequences.

Given:

1. $a_1, a_2, a_3, \dots$ are in A.P.
2. $h_1, h_2, h_3, \dots$ are in H.P.
3. $a_1 = 2 = h_1$
4. $a_{30} = 25 = h_{30}$

We need to find the value of $(a_{24} + a_{14}h_{17})$.

A key property for A.P. and H.P. is that if $a_1 = h_1$ and $a_N = h_N$, then for any $n$, $a_n h_{N+1-n} = a_1 a_N$. In this problem, $N=30$. So, $a_n h_{30+1-n} = a_n h_{31-n} = a_1 a_{30}$.

Substituting the given values: $a_n h_{31-n} = 2 \times 25 = 50$.

For the term $a_{14}h_{17}$, the sum of the indices is $14+17=31$. This fits the property $a_n h_{31-n}$.

So, $a_{14}h_{17} = a_1 a_{30} = 50$.

The term $a_{24}$ does not simplify to an integer, so it is likely a flawed question or a typo. Assuming the most likely intended answer from the options based on the property, the answer would be 50.