Question

The sum of the 1st three terms of an AP is 42 and the product of the 1st and 3rd term is 52. Find the 1st term and common difference

Answer 1

Answer :-

d = ± 12

a = 2 , 26

Given :-

$S_3 = 42$

$a \times a_3 = 52$

To find :-

The 1st term and common difference.

Solution:-

Let the first term of AP be a and common difference be d.

A/Q

$S_3 = 42$

→$\dfrac{3}{2} [2a + (3-1)d ] = 42$

→$\dfrac{3}{2}[2a + 2d ] = 42$

→$3(2a + 2d) = 42 \times 2$

→$6a +6d = 84$

Dividing both side by 6,

→$\dfrac{6a}{6}+ \dfrac{6d}{6}= \dfrac{84}{6}$

→$a + d = 14$

→$a = 14 -d ----1$

$a \times a_3 = 52$

→$a ( a +2d ) = 52$

→$(14 -d) ( 14-d + 2d) = 52$

→$(14-d) (14+d) = 52$

→$196 -d^2 = 52$

→$196 -52 = d^2$

→$144 = d^2$

→$d = \sqrt{144}$

→$d = \pm12$

Put the value of d in eq.1

If d = 12

a = 14 - d

a = 14 - 12

a = 2

If d = -12

a = 14 - ( -12)

a = 14 + 12

a = 26

hence,

The first term is 2 and 26 and common difference is ± 12.

Answer 2

GIVEN :

Let the terms be a - d, a and a + d

Sum of 1st three terms of an AP = 42

a - d + a + a + d = 42

=> 3a = 42

=> a = 42/3

=> a = 14

Product of 1st and 3rd term is 52

(a - d) (a + d) = 52

a² - d² = 52 [ (a - b) (a + b) = a² - b² ]

(14)² - d² = 52

196 - d² = 52

196 - 52 = d²

=> d² = 144

=> d = √144

=> d = ±12

Common Difference = ±12

First term :

If d is -12

= a - d

= 14 - (-12)

= 26

If d is +12

= a - d

= 14 - (12)

= 2

Therefore, 2 or 26 is first term and ±12 is difference of AP respectively.