Let be the term of an A.P. If for some m, and , then is equal to [2025]

Question

Let $T_{r}$ be the $r^{th}$ term of an A.P. If for some m, $T_{m} = \frac{1}{25}, T_{25} = \frac{1}{20}$ and $20 \sum_{r = 1}^{25} T_{r} = 13$ , then $5 m \sum_{r = m}^{2 m} T_{r}$ is equal to [2025]

Smart Achievers · Accepted Answer

(2)

We have, $T_{m} = \frac{1}{25}, T_{25} = \frac{1}{20}$ and $20 \sum_{r = 1}^{25} T_{r} = 13$

$\Rightarrow T_{m} = a + (m - 1) d = \frac{1}{25}$ ... (i)

and $T_{25} = a + 24 d = \frac{1}{20}$ ... (ii)

Also, $20 \sum_{r = 1}^{25} T_{r} = 13$

$\Rightarrow 20 [\frac{25}{2} (2 a + 24 d)] = 13$

$\Rightarrow 20 [\frac{25}{2} (a + \frac{1}{20})] = 13$ (Using eqn. (ii))

$\Rightarrow a = \frac{1}{500}, d = \frac{1}{500}$

Using (i), we get m = 20

Now, $5 m \sum_{r = m}^{2 m} T_{r} = 5 \times 20 \sum_{r = 20}^{40} T_{r} = 126$ .

Solution

Q.
Let $T_{r}$ be the $r^{th}$ term of an A.P. If for some m, $T_{m} = \frac{1}{25}, T_{25} = \frac{1}{20}$ and $20 \sum_{r = 1}^{25} T_{r} = 13$ , then $5 m \sum_{r = m}^{2 m} T_{r}$ is equal to [2025]

1 112

2 126

3 142

4 98

Want Us to Email you About Special Offers & Updates?