Let an be the n th term of an A.P. If Sn=a1+a2+⋯+an=700 for some n, a6=7 and S7=7, then an is equal to:
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Step-by-Step Solution
Step 1: Formulate equations from given information
We are given an arithmetic progression (A.P.). The general formula for the k -th term of an A.P. is ak=a+(k−1)d, where a is the first term and d is the common difference. The sum of the first k terms of an A.P. is given by Sk=2k(2a+(k−1)d). We will use these formulas to set up equations based on the given conditions.
Step 2: Solve for 'a' and 'd' using given conditions
We are given a6=7 and S7=7. Using the formulas from the previous step, we can write these as two linear equations in terms of a and d. From a6=7, we get a+5d=7. From S7=7, we get 27(2a+(7−1)d)=7, which simplifies to 7(a+3d)=7, and further to a+3d=1. Now we have a system of two linear equations.
Step 3: Find the common difference 'd' and first term 'a'
To find d, we subtract equation (2) from equation (1): (a+5d)−(a+3d)=7−1, which simplifies to 2d=6, so d=3. Now substitute d=3 into equation (2): a+3(3)=1, which gives a+9=1, so a=−8.
Step 4: Find 'n' using the sum formula
We are given Sn=700. Substitute the values of a=−8 and d=3 into the sum formula: Sn=2n(2(−8)+(n−1)3)=700. This simplifies to a quadratic equation: 3n2−19n−1400=0.
Step 5: Solve the quadratic equation for 'n'
We solve the quadratic equation 3n2−19n−1400=0 using the quadratic formula n=2a−b±b2−4ac. Substituting the coefficients, we get n=2(3)19±(−19)2−4(3)(−1400). This simplifies to n=619±361+16800=619±17161=619±131. Since n must be a positive integer, we take the positive root: n=619+131=6150=25. The negative root is rejected.
Step 6: Calculate the 'n'th term, an
Now that we have a=−8, d=3, and n=25, we can find the n -th term, an, using the formula an=a+(n−1)d. Substituting the values, we get a25=−8+(25−1)3=−8+24×3=−8+72=64.