The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of the first sixteen terms of the AP.
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Step-by-Step Solution
Step 1: Define the terms of an AP
Let a be the first term and d be the common difference of the AP. The n -th term of an AP is given by the formula an=a+(n−1)d. We will use this to express the third and seventh terms.
Step 2: Express the given conditions using the AP formula
We write the third term as a3=a+2d and the seventh term as a7=a+6d. The problem states that their sum is 6 and their product is 8. We will form equations based on these conditions.
Step 3: Formulate equations from the given conditions
From the given information, we form two equations. Equation 1 represents the sum of the third and seventh terms, and Equation 2 represents their product. We now have a system of two equations with two variables, a and d.
Step 4: Simplify Equation 1
Simplifying Equation 1, we combine like terms: a+2d+a+6d=2a+8d=6. Dividing by 2, we get a+4d=3. This simplified equation will be easier to use for substitution.
Step 5: Express 'a' in terms of 'd' from Equation 3
From Equation 3, we can express a in terms of d. This allows us to substitute a into Equation 2, thereby reducing it to a single variable equation in d.
Step 6: Substitute 'a' into Equation 2
Substitute a=3−4d into Equation 2 (a+2d)(a+6d)=8. This step transforms Equation 2 into an equation solely involving d, which we can then solve.
Step 7: Simplify and solve for 'd'
Simplify the substituted equation using the difference of squares formula, (x−y)(x+y)=x2−y2. This leads to a quadratic equation in d, for which we find two possible values for d: 1/2 and −1/2.
Step 8: Case 1: Calculate 'a' when d=1/2
Using Equation 3 (a=3−4d), we substitute d=1/2 to find the corresponding value for a. In this case, a=1. This gives us one possible AP.
Step 9: Calculate sum of first 16 terms for Case 1
The sum of the first n terms of an AP is given by the formula Sn=2n[2a+(n−1)d]. For n=16, a=1, and d=1/2, the sum S16 is calculated as 76.
Step 10: Case 2: Calculate 'a' when d=−1/2
Now we consider the second possible value for d. Substituting d=−1/2 into Equation 3 (a=3−4d), we find a=5. This gives us another possible AP.
Step 11: Calculate sum of first 16 terms for Case 2
Again, using the sum formula Sn=2n[2a+(n−1)d], with n=16, a=5, and d=−1/2, we find S16=20. Both values of d lead to valid APs.