एक श्रृंखला समान्तर श्रेणी में है, यदि अंतिम पद 21 है और सभी 10 पदों का योग 120 है, तब पहले तीन पदों का योग ज्ञात कीजिए।

The sum of $n$ terms is given by $S_n = \frac{n}{2}(a_1 + a_n)$.
Given $S_{10} = 120$, $n = 10$, and $a_{10} = 21$.

Divide both sides by 5:

Subtract 21 from both sides:

The $n$-th term is given by $a_n = a_1 + (n-1)d$.
Given $a_{10} = 21$, $a_1 = 3$, and $n = 10$.

Subtract 3 from both sides:

Divide both sides by 9:

The first term is $a_1 = 3$.
The second term is $a_2 = a_1 + d = 3 + 2 = 5$.
The third term is $a_3 = a_2 + d = 5 + 2 = 7$.

The sum of the first three terms is $S_3 = a_1 + a_2 + a_3$.

The sum of the first three terms is $\text{15}$.