Question

Hola!!!

The sum of first three terms of an Arithmetic Progression is 42 and the product of the first and third term is 53. Find the first term and common difference.​

Answer 1

Let a-d , a , a+d are first three

terms in an A.P

According to the problem given,

i ) Sum of three terms = 42

=> a-d+a+a+d=42

=> 3a = 42

=> a = 42/3

=> a = 14

ii ) Product of the first and third term is 52

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

=> a² - d² = 52

=> 14² - d² = 52

=> 196 - d² = 52

=> -d² = 52 - 196

=> -d² = -144

=> d² = 144

=> d = ±√144
=> d = ± 12

Therefore,

first term = a-d = 14±√12
= ( 14 +12 ) or (14-12)
= 26 or 2

common difference = a2-a1

= a - (a-d)

= a - a + d

= ± 12

••••

Answer 2

Question:The sum of first three terms of an Arithmetic Progression is 42 and the product of the first and third term is 52. To find the first term and common difference.

let us assume that the numbers are a-d,a,a+d
,it is convenient to take three numbers like that

So, A.T.Q

The sum of first three terms of an Arithmetic Progression is 42 ,So

$a - d + a + a + d = 42 \\ \\ 3a = 42 \\ \\ a = \frac{42}{3} \\ \\ a = 14$
the product of the first and third term is 52,so

$(a + d)(a - d) = 52 \\ \\ {a}^{2} - {d}^{2} = 52$
put the value of a here

$( {14)}^{2} - {d}^{2} = 52 \\ \\ 196 - {d}^{2} = 52 \\ \\ {d}^{2} = 196 - 52 \\ \\ {d}^{2} = 144 \\ \\ d = \sqrt{144} \\ \\ d = + - 12$
so, d can be 12 or -12.

Thus first term of AP; a=14
common difference d= 12

Verification:

with a=14,d=12

Numbers are 14-12,14,14+12

2,14,26

Sum=2+14+26=42

2(26)=52

with a=14,d=-12

Numbers are 14+12,14,14-2

26,14,2

Sum=2+14+26=42

2(26)=52

Hence proved.

Hope it helps you.