Let a1,a2,a3,… be a G.P. of increasing positive terms. If a1a5=28 and a2+a4=29, then a6 is equal to:
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Step-by-Step Solution
Step 1: Express terms in terms of a1 and r
In a geometric progression (G.P.), each term can be expressed using the first term a1 and the common ratio r. The n -th term is given by the formula an=a1rn−1. Since the terms are increasing and positive, the common ratio r must be greater than 1.
Step 2: Formulate equations from given conditions
Using the general formula for the n -th term, we can rewrite the given conditions. The first condition a1a5=28 becomes a1(a1r4)=28, which simplifies to a12r4=28. The second condition a2+a4=29 becomes a1r+a1r3=29, which can be factored as a1r(1+r2)=29.
Step 3: Solve for a1 and r
From equation (1), a12r4=28, taking the square root gives a1r2=28=27. We can express a1 as a1=r227. Substitute this into equation (2): r227r(1+r2)=29. This simplifies to 27r1+r2=29, which leads to the quadratic equation 27r2−29r+27=0. Solving this quadratic equation for r using the quadratic formula yields two possible values for r: 27 and 271. Since the G.P. has increasing terms, r must be greater than 1. Therefore, r=27. Substituting this value back into the expression for a1 gives a1=147.
Step 4: Calculate a6
Now that we have a1=147 and r=27, we can find the 6th term, a6. Using the formula a6=a1r5, we substitute the values: a6=(147)(27)5. This simplifies to a6=147⋅32⋅(7)5=147⋅32⋅497. Multiplying the terms, we get a6=147⋅32⋅49=21⋅32⋅49=16⋅49=784.