Maths question from Panchayat Secretary (VLW) exam, 2025 by JKSSB
An arithmetic progression with first term 19 has sum of its first 20 terms as 2280. What will be the sum of its first 30 terms if the common difference is doubled?
Last updated May 30, 2026
Correct Answer:
Option C —
8685
Step-by-Step Explanation
Step 1: Find the original common difference
An arithmetic progression is a sequence of numbers where the difference between consecutive terms is constant. We are given:
The first term is 19.
The sum of the first 20 terms is 2280.
The standard formula for the sum of an arithmetic progression can be thought of as: the number of terms divided by 2, multiplied by the sum of twice the first term and the number of terms minus 1 times the common difference.
Applying this to our first scenario:
20 terms divided by 2 is 10.
Twice the first term is 38 (2 times 19).
The number of terms minus 1 is 19 (20 minus 1).
So, 10 multiplied by the quantity (38 plus 19 times the common difference) equals 2280.
If we divide both sides by 10, we get:
38 plus 19 times the common difference equals 2288 divided by 10, which is 228.
Subtract 38 from both sides:
19 times the common difference equals 190.
Divide by 19:
The original common difference is 10.
Step 2: Calculate the new sum with the doubled common difference
Now, the problem states the common difference is doubled, and we need the sum of the first 30 terms:
The first term remains 19.
The new common difference is 20 (2 times 10).
The number of terms is now 30.
Let us use the sum logic again:
30 terms divided by 2 is 15.
Twice the first term is still 38.
The new number of terms minus 1 is 29 (30 minus 1).
29 times the new common difference (20) equals 580.
Now, add twice the first term and this new value together:
38 plus 580 equals 618.
Finally, multiply this result by 15 (which was 30 terms divided by 2):
15 times 618 equals 8685.
Answer verified by Quintessence Classes faculty — Karan Nagar, Srinagar.
