Question

║⊕QUESTION⊕║
“Wherever there is a number, there is beauty.” -Proclus
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CLASS 11
SEQUENCES AND SERIES
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Find three numbers in AP whose sum is 42 and whose product is 2618.

Answer 1

Answer:

11,14,17

Step-by-step explanation:

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Let the numbers be a-d, a, a+d

Then,

3a = 42

a = 14

Also,

(a-d)(a+d)a = 2618

a(a^2 - d^2) = 2618

d^2 = 196 - (2618/14)

= 196 - 187

= 9

d = 3

Numbers : 11,14,17

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Answer 2

Answer:

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let a - d, a,a +d are the three consecutive terms in arithmatic progression.

according to question

sum of these terms = 42

(a - d) + (a )+ (a + d)= 42

or, 3a = 42

hence. a= 14 ..............(1)

and also given that ,

product of these terms = 2618

(a - d)(a)(a+ d)= 2618

( a^2 - d^2)=2618/a

from equation (1) a= 14

196 - d^2 = 187

so ,d^2= 196-187

d= √9

d= 3 ,

therefore,

first term = a - d= 14-3= 11

second term = a= 14

third term = a+ d= 14+3= 17

therefore required AP is

11,14,17..

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