We know that the n -th term of an AP is given by the formula a_n = a + (n - 1)d, where a is the first term and d is the common difference. For the 5th term, substituting n = 5 gives a + 4d = 18. For the 10th term, substituting n = 10 gives a + 9d = 30.

We know that the n -th term of an AP is given by the formula a_n = a + (n - 1)d, where a is the first term and d is the common difference. For the 5th term, substituting n = 5 gives a + 4d = 18. For the 10th term, substituting n = 10 gives a + 9d = 30.

The sum of the first n terms is given by S_n = (n)/(2)[2a + (n - 1)d]. For n = 25, substitute a = (42)/(5) and d = (12)/(5). Inside the brackets, 2((42)/(5)) + 24((12)/(5)) = (84 + 288)/(5) = (372)/(5) = 74.4. Then, S_25 = (25)/(2) × 74.4 = 930.

The sum of the first n terms is given by S_n = (n)/(2)[2a + (n - 1)d]. For n = 25, substitute a = (42)/(5) and d = (12)/(5). Inside the brackets, 2((42)/(5)) + 24((12)/(5)) = (84 + 288)/(5) = (372)/(5) = 74.4. Then, S_25 = (25)/(2) × 74.4 = 930.

the 5th term of an ap is 18 and 10 is 30 find the sum of the first 25 terms

Answer: $930$

Step-by-step solution

Step 1: Set up equations using the nth term formula

We know that the $n$ -th term of an AP is given by the formula $a_n = a + (n - 1)d$ , where $a$ is the first term and $d$ is the common difference. For the $5\text{th}$ term, substituting $n = 5$ gives $a + 4d = 18$ . For the $10\text{th}$ term, substituting $n = 10$ gives $a + 9d = 30$ .

Step 2: Solve for the common difference d

Subtract the first equation from the second equation to eliminate $a$ : $(a + 9d) - (a + 4d) = 30 - 18$ , which simplifies to $5d = 12$ . Dividing both sides by $5$ , we find that $d = 2.4$ (or $\frac{12}{5}$ ).

Step 3: Find the first term a

Substitute the value of $d = \frac{12}{5}$ back into the first equation $a + 4d = 18$ . This gives $a + 4\left(\frac{12}{5}\right) = 18$ , so $a = 18 - 9.6 = 8.4$ (or $\frac{42}{5}$ ).

Step 4: Calculate the sum of the first 25 terms

The sum of the first $n$ terms is given by $S_n = \frac{n}{2}[2a + (n - 1)d]$ . For $n = 25$ , substitute $a = \frac{42}{5}$ and $d = \frac{12}{5}$ . Inside the brackets, $2\left(\frac{42}{5}\right) + 24\left(\frac{12}{5}\right) = \frac{84 + 288}{5} = \frac{372}{5} = 74.4$ . Then, $S_{25} = \frac{25}{2} \times 74.4 = 930$ .

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