Question

Find four consecutive terms in an A.P such that the sum of middle terms is 26 and the product of the two end terms is 88.

Answer 1

let them be x-3d x-d x+d x+3d therefore x+x =26 x=13 also x2 - (3d)2 = 88 solve the rest

Answer 2

Let the four consecutive terms of A.P are a, a+d, a+2d, a+3d ,
where  a is 1st term and d is the common difference.
A/Q,
Sum of two middle terms = 26
⇒ a+d + a+2d = 26
⇒ 2a+3d = 26
3d = 26 - 2a
⇒  d = (26-2a)/3    ---------------(i)
and,
Product of two end terms = 88
⇒ (a).(a+3d) = 88
⇒  a² +3ad = 88
⇒  a² + 3a(26-2a)/3 = 88
⇒  a² + a(26-2a) = 88
⇒  a² + 26a -2a² = 88
⇒  -a² + 26a -88 = 0
⇒  a² - 26a + 88 = 0
⇒  a² - 22a - 4a + 88 = 0
⇒  a(a - 22) - 4(a - 22) = 0
⇒  (a - 4)(a - 22) = 0
⇒  a - 4 = 0
        a =  4
put the value of a in Equation (i)
       d = (26 - 2*4)/3 = (26 - 8)/3
⇒    d = 18/3 = 6
        d = 6
Therefore four consecutive terms of A.P are:-
4, 4+6, 4+2*6 and 4+3*6
i.e,
4, 10, 16 and 22


or,
⇒   a - 22 = 0 ⇒ a = 22
put the value of a in Equation (i)
    d = (26  - 2*22)÷3
⇒ d = (26 - 44)/3
⇒ d = -18/3 = -6
Therefore four consecutive terms of A.P are:-
22, 22-6, 22-2*6 and  22-3*6
i.e
22, 16, 10 and 4.