Question

Four numbers are in arithmetic progression. The sum of first and last term is 8 and the product of both middle terms is 15. The least number of the series is.

• A. 4
• B. 3
• C. 2
• D. 1

Answer 1

Answer: (D)

1

Answer 2

Let $A _ { 1 } , A _ { 2 } , A _ { 3 }$ and $A _ { 4 }$ are four numbers in A.P.

$A _ { 1 } + A _ { 4 } = 8$ …..(i) and $A _ { 2 } \cdot A _ { 3 } = 15$ …..(ii)

The sum of terms equidistant from the beginning and end is constant and is equal to sum of first and last terms.

Hence, $A _ { 2 } + A _ { 3 } = A _ { 1 } + A _ { 4 } = 8$ …..(iii)

From (ii) and (iii),

$A _ { 2 } + \frac { 15 } { A _ { 2 } } = 8$ ⇒ $A _ { 2 } ^ { 2 } - 8 A _ { 2 } + 15 = 0$

$A _ { 2 } = 3$ or 5 and $A _ { 3 } = 5$ or 3.

As we know, $A _ { 2 } = \frac { A _ { 1 } + A _ { 3 } } { 2 }$ ⇒ $A _ { 1 } = 2 A _ { 2 } - A _ { 3 }$

⇒ $A _ { 1 } = 2 \times 3 - 5 = 1$ and $A _ { 4 } = 8 - A _ { 1 } = 7$

Hence the series is, 1, 3, 5, 7.

So that least number of series is 1.