3 a) Simplify (2x+14)/(2x2-98). b) Given that the third and sixth terms of a geometric progression are 3/4 and 3/32 respectively, find the i) first term and the common ratio, ii) nth term, iii) sum to infinity of the progression.

a) Step 1: Factor out common terms from the numerator and denominator. The numerator is 2x+14. We can factor out 2: 2(x+7). The denominator is 2x^2-98. We can factor out 2: 2(x^2-49). Step 2: Apply the difference of squares formula to the denominator. The expression x^2-49 is in the form a^2-b^2 = (a-b)(a+b), where a=x and b=7. So, x^2-49 = (x-7)(x+7). Step 3: Substitute the factored forms back into the fraction and simplify. (2x+14)/(2x^2-98) = (2(x+7))/(2(x^2-49)) = (2(x+7))/(2(x-7)(x+7)) Cancel out the common factors of 2 and (x+7). (1)/(x-7) The simplified expression is (1)/(x-7). b) Given that the third term (T_3) of a geometric progression is (3)/(4) and the sixth term (T_6) is (3)/(32). The formula for the n^th term of a geometric progression is T_n = ar^n-1, where a is the first term and r is the common ratio. i) First term and the common ratio Step 1: Write equations for T_3 and T_6. For T_3: ar^3-1 = ar^2 = (3)/(4) (Equation 1) For T_6: ar^6-1 = ar^5 = (3)/(32) (Equation 2) Step 2: Divide Equation 2 by Equation 1 to find the common ratio r. (ar^5)/(ar^2) = (3)/(32)(3)/(4) r^3 = (3)/(32) × (4)/(3) r^3 = (4)/(32) r^3 = (1)/(8) Step 3: Solve for r. r = 3/(8) r = (1)/(2) Step 4: Substitute the value of r into Equation 1 to find the first term a. a((1)/(2))^2 = (3)/(4) a((1)/(4)) = (3)/(4) a = 3 The first term is a=3 and the common ratio is r=(1)/(2). ii) n^th term Step 1: Use the formula for the n^th term, T_n = ar^n-1, with a=3 and r=(1)/(2). T_n = 3((1)/(2))^n-1 The n^th term is 3((1)/(2))^n-1. iii) Sum to infinity of the progression Step 1: Use the formula for the sum to infinity of a geometric progression, S_ = (a)/(1-r). This formula is valid when |r| < 1. Since r=(1)/(2), |r|=(1)/(2) < 1, so the sum to infinity exists. Step 2: Substitute a=3 and r=(1)/(2) into the formula. S_ = (3)/(1 - 1)2 S_ = (3)/(1)2 S_ = 3 × 2 S_ = 6 The sum to infinity of the progression is 6. That's 2 down. 3 left today — send the next one.