Question

In an ap the sum of second and third term is 22 the product of first anhd fourth term is 85 find AP​

Answer 1

Given :

• Sum of 2nd & 3rd term = 22
• Product of 1st & 4th term = 85

To Find :

• Arithmetic progression (A.P)

Formula used :

• Tn = a +(n-1)d

▪️Tn is the nth term of the A.P

▪️a is the first term.

▪️n is the number of terms.

▪️d is common difference

Solution :

According to the question,

Case I,

T2 + T3 = 22

→a +(2-1)d + a + (3-1) = 22

→a + d+ a + 2d = 22

→2a + 3d = 22

Case II,

T1 × T4 = 85

→[a + (1-1)d] × [a +(4-1)d] = 85

→[a]×[a+3d] = 85

→a ² + 3ad = 85

From Case I,

2a + 3d = 22

3d = 22-2a

3d = 2(11-a)

$d = \frac{2}{3} (11 - a)$

Using this value in case II we get,

a ² + 3ad = 85

${a}^{2} + 3a \times\frac{2}{3} (11 - a) = 85 \\ {a}^{2} + 2a(11 - a) = 85 \\ {a}^{2} + 22a - 2 {a}^{2} = 85 \\ 22a - {a}^{2} = 85 \\ - ( - {a}^{2} + 22a - 85) = 0 \\ {a}^{2} - 22a + 85 = 0$

On solving this quadratic equation,

→a ² - 17a - 5a +85 = 0

→a(a-17) - 5(a-17) = 0

→(a - 5)(a-17) = 0

So, a = 5 or 17.

Now, when a = 5

$d = \frac{2}{ 3} (11 - 5) \\ d = \frac{2}{3} \times 6 = 4$

Hence, the A.P we get as : 5,9,13,17.....so on.

And, when a = 17

$d = \frac{2}{3} (11 - 17) \\ d = \frac{2}{3} \times - 6 = - 4$

Hence, the A.P we get as : 17,13,9,5....so on.