Sequence and Series – Arithmetic Progression – JEE Main 29 Jan 2025 Shift 1

Solution:

Let the A.P. be $a, a+d, a+2d, \dots$. All terms are positive integers, so $a \in Z^+$ and $d \in Z$.

Given sum of first 3 terms $S_3 = 54$.

$\frac{3}{2}[2a + (3-1)d] = 54$

$3(a+d) = 54 \Rightarrow a+d = 18 \Rightarrow a = 18-d$.

Since $a$ is a positive integer, $18-d > 0 \Rightarrow d < 18$.
Given sum of first 20 terms $S_{20}$ lies between 1600 and 1800.

$S_{20} = \frac{20}{2}[2a + (20-1)d] = 10[2a + 19d]$.

Substitute $a = 18-d$:

$S_{20} = 10[2(18-d) + 19d] = 10[36 – 2d + 19d] = 10[36 + 17d]$.

Inequality: $1600 < 10(36 + 17d) < 1800$.
Divide by 10:

$160 < 36 + 17d < 180$.
Subtract 36:

$124 < 17d < 144$.
Divide by 17:

$\frac{124}{17} < d < \frac{144}{17}$.
$7.29 < d < 8.47$.
Since $d$ must be an integer (as terms are integers), the only possible value is $d = 8$.

Find $a$: $a = 18 – 8 = 10$.

We need to find the $11^{\text{th}}$ term, $a_{11}$.

$a_{11} = a + 10d = 10 + 10(8) = 10 + 80 = 90$.

Ans. (3)