Sequence & Series – Arithmetic Progression – JEE Main 28 Jan 2025 Shift 1

Solution:

Let the first term be $a$ and the common difference be $d$.
Given:
1) $T_m = a + (m-1)d = \frac{1}{25}$ … (i)
2) $T_{25} = a + 24d = \frac{1}{20}$ … (ii)

Subtract (i) from (ii):
$(25-m)d = \frac{1}{20} – \frac{1}{25} = \frac{5 – 4}{100} = \frac{1}{100}$
$\Rightarrow d = \frac{1}{100(25-m)}$

Also given sum condition:
$20 \sum_{r=1}^{25} T_r = 13$
$20 \times \frac{25}{2} [2a + 24d] = 13$
$250 [2a + 24d] = 13$
From (ii), $a + 24d = 1/20 \Rightarrow 2a + 48d = 1/10 \Rightarrow 2a + 24d = 1/10 – 24d$.
Alternatively, note that $T_1 + T_{25} = 2a + 24d$.
$20 \times \frac{25}{2} (T_1 + T_{25}) = 13$
$250 (T_1 + \frac{1}{20}) = 13$
$250 T_1 + \frac{250}{20} = 13$
$250 T_1 + 12.5 = 13 \Rightarrow 250 T_1 = 0.5 \Rightarrow T_1 = \frac{1}{500}$.
So, $a = \frac{1}{500}$.

Substitute $a$ in (ii):
$\frac{1}{500} + 24d = \frac{1}{20}$
$24d = \frac{1}{20} – \frac{1}{500} = \frac{25-1}{500} = \frac{24}{500}$
$\Rightarrow d = \frac{1}{500}$.

Now find $m$ using (i):
$T_m = \frac{1}{500} + (m-1)\frac{1}{500} = \frac{1}{25}$
$\frac{1 + m – 1}{500} = \frac{1}{25}$
$\frac{m}{500} = \frac{1}{25} \Rightarrow m = 20$.

We need to find $S_{req} = 5m \sum_{r=m}^{2m} T_r$.
Substitute $m=20$:
$S_{req} = 100 \sum_{r=20}^{40} T_r$.
The sum is of terms from $T_{20}$ to $T_{40}$ (Total terms = $40-20+1 = 21$).
Sum $= \frac{n}{2} [T_{first} + T_{last}] = \frac{21}{2} [T_{20} + T_{40}]$.

$T_{20} = a + 19d = \frac{1}{500} + \frac{19}{500} = \frac{20}{500}$.
$T_{40} = a + 39d = \frac{1}{500} + \frac{39}{500} = \frac{40}{500}$.
Sum $= \frac{21}{2} \left[ \frac{60}{500} \right] = \frac{21}{2} \times \frac{6}{50} = \frac{21 \times 3}{50} = \frac{63}{50}$.

$S_{req} = 100 \times \frac{63}{50} = 2 \times 63 = 126$.

Ans. (2)