Question

(i) The sum of the first 41 terms of an A.P. is 5125. Complete the following

activity to find the 21st term.

Sn =S41 =5125,
​

Answer 1

Given

Sum of the first 41 terms of an A.P. is 5125

To find

The 21st term.

Solution

We can simply solve this mathematical problem using the following mathematical process.

∴ $S_{n}$ = $\frac{h}{2}$ [ 2a + (n - 1) d ]

∴ $S_{n}$ = $\frac{41}{2}$ [ 2a + (41 - 10 d]

5125 = $\frac{41}{2}$( 2a + 40 d)

∴ 5125 = 41 × (a + 40d)

∴  a + 20d = $\frac{5125}{41}$

a + 20d = 125 ·········· (1)

Now, 21st term is $t_{21}$

∴ $t_{n}$ = a (n-1) d

$t_{21}$ = a + (21 - 1) d

$t_{21}$ = a + 20d

$t_{21}$ = 125

Hence, the value of  $t_{21}$ is 125

Answer 2

Answer:

The 21 st term of the AP will be 125.

Step-by-step explanation:

Given: The sum of the first 41 terms of an A.P. is 5125, a is the first term and d is the Common Difference of the Arithmetic progression

We have to find the 21st term of the A.P.

We are solving in the following way:

We have,

The sum of the first 41 terms of an A.P. is 5125.

Let's assume, a is the first term and d is the Common Difference of the Arithmetic progression

As we know, the sum of the first n terms of the AP is calculated by the formula:

$\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$

Also, the $n^t^h$term of the AP is calculated by the formula:$a+(n-1) d$

Now it is given that the sum of the first 41 terms is 5125.

Then,

$\begin{array}{l}\Longrightarrow \frac{41}{2}[2 a+(41-1) d]=5125 \\\\\Longrightarrow \frac{41}{2}[2 a+40 d]=5125 \\\\\Longrightarrow \frac{41}{2} \times 2 \times[a+20 d]=5125 \\\\\Longrightarrow 41 \times[a+20 d]=5125 \\\\\Longrightarrow[a+20 d]=125\end{array}$

The 21 st term of the AP will be:

$\begin{array}{l}=\mathrm{a}+(21-1) \mathrm{d} \\=\mathrm{a}+20 \mathrm{~d} \\=125\end{array}$

Hence, The 21 st term of the AP will be 125.