In an arithmetic progression, if S40=1030 and S12=57, then S30−S10 is equal to:
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Step-by-Step Solution
Step 1: Define the sum of an arithmetic progression
The sum of the first n terms of an arithmetic progression (AP) is given by the formula Sn=2n[2a+(n−1)d], where a is the first term and d is the common difference.
Step 2: Formulate equations from given information
We are given S40=1030 and S12=57. Substituting these values into the sum formula, we get two linear equations in terms of a and d. Simplifying these equations gives us 2a+39d=51.5 and 2a+11d=9.5.
Step 3: Solve for 'a' and 'd'
To find the values of a and d, we subtract equation (2) from equation (1). This eliminates 2a, allowing us to solve for d. Once d is found, we substitute it back into equation (2) to solve for a. We find that d=1.5 and a=−3.5.
Step 4: Calculate S30 and S10
Now that we have the values for a and d, we can calculate S30 and S10 using the sum formula. Substituting n=30, a=−3.5, and d=1.5 gives S30=547.5. Similarly, substituting n=10, a=−3.5, and d=1.5 gives S10=32.5.
Step 5: Find the difference S30−S10
Finally, we subtract S10 from S30 to find the required value. The difference is 547.5−32.5=515.